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two students on a balcony 18.8m above the street. one student throws a ball vertically DOWNWARD 13.1 m/s; at the same instant, the onther st

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two students on a balcony 18.8m above the street. one student throws a ball vertically DOWNWARD 13.1 m/s; at the same instant, the onther student throws a ball vertically UPWARD at the same speed. what is the volcity of the first ball as it strikes the ground? assume it is in the posotive direction... what is the volcity of the second ball as it strikes the ground? what is the difference in the time the b***s spend in the air? how far apart are the b***s 0.964s after they are thrown?

Sep 16, 2014

#7
+20850
+5

two students on a balcony 18.8m above the street. one student throws a ball vertically DOWNWARD 13.1 m/s; at the same instant, the onther student throws a ball vertically UPWARD at the same speed.

I. what is the volcity of the first ball as it strikes the ground? assume it is in the posotive direction... what is the volcity of the second ball as it strikes the ground?

$$g=9.81 \frac{m}{s^2} \quad v_0 = 13.1\frac{m}{s} \quad h=18.8\;m$$

$$\small{ \text{Way vertically downward: } s_1 = \frac{g}{2}t^2 \quad \text{and way thrown vertically downward or upward: } {s_2=v_0t\quad }$$

$$\small{ way \;first\; ball\; s_I:\boxed{s_I= s_1+s_2 = h} \quad way \;second\; ball s_{II}:\boxed{s_{II}= s_1-s_2 = h} }$$

$$\begin{array}{rcl|rcl} \small{ \text{time first ball:} \ t_1 } &&& \small{ \text{time second ball:} \ t_2 }\\ s_I= \frac{g}{2}t_1^2+v_0t_1 & = & h & s_{II}=\frac{g}{2}t_2^2-v_0t_2 &=&h\\ \frac{g}{2}t_1^2+v_0t_1 -h& = & 0 & \frac{g}{2}t_2^2-v_0t_2 -h&=& 0 \\ t_1>0: \quad t_1 &=& \frac{-v_0+\sqrt{v_0^2+2gh} }{g} &t_2>0: \quad t_2 &=& \frac{v_0+\sqrt{v_0^2+2gh} }{g} \\ \end{array}$$

$$\begin{array}{rcl|rcl} \small{ \text{final speed first ball:} \ v_1 } &&& \small{ \text{final speed second ball:} \ v_2 }\\ v_1&=& gt_1+v_0 & v_2 &=& gt_2-v_0 \\ v_1 & = & -v_0 + \sqrt{v_0^2+2gh} + v_0 & v_2 &=& v_0 + \sqrt{v_0^2+2gh} - v_0 \\ \end{array}$$

$$\boxed{v_1=v_2=\sqrt{v_0^2+2gh} } = \sqrt{13.1^2+2*9.81*18.8} = 23.2479246386\ \frac{m}{s}$$

II. what is the difference in the time the b***s spend in the air?

$$\begin{array}{rcl} \Delta t & = & t_2-t_1 \\ &=& \frac{(v_0 + \sqrt{v_0^2+2gh})}{g} - \frac{( -v_0 + \sqrt{v_0^2+2gh} ) } {g}\\ &=& \frac{2v_0}{g} \end{array}$$

$$\boxed{\Delta t=\frac{2v_0}{g} } = \frac{2*13.1}{9.81} = 2.67074413863\ s$$

III. how far apart are the b***s 0.964s after they are thrown?

$$\begin{array}{rcl} \Delta s & = & s_I-s_{II} \\ &=& (s_1 + s_2 ) - ( s_1 - s_2 )\\ &=&2s_2 \\ &=&2v_0t \end{array}$$

$$\boxed{ \Delta s=2v_0t } = 2*13.1*0.964 = 25.2568\ m$$

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Sep 17, 2014

#1
+2353
+5

Hi there!

You've got a fairly challenging question there, but I can help you out .

Both b***s have potential energy since they are at a balcony at a height of 18.8 meters.

The formula for potential energy is given by

$$E_p = mgh$$

where Ep is the potential energy, m is the mass, g is the acceleration of gravity and h is the height.

Generally g is approximated to be about 9.8 (this differs across the globe).

Therefore $$E_p = m*9.8*18.8 = 184.24*m$$

In addition to that we know that the first ball is thrown downwards with a force of 13.1 m/s

The formula for kinetic energy is given by

$$E_k = \frac{1}{2}mv^2$$

where Ek is the kinetic energy, m is the mass and v is the velocity (speed).

Therefore we know this throw has given an additional energy of.

$$E_k = \frac{1}{2}m(13.1)^2 = 85.805*m$$

So the total energy when the ball is thrown downwards is given by

$$E_t = E_k + E_p = 184.24*m+85.805*m = 270.045*m$$

Now by the first law of thermodynamics we know that energy always remains constant.

If we neglect air resistance, all energy should have been transformed into kinetic energy, a fraction of a milisecond before it hits the ground.

Therefore

$$\begin{array}{lcl} 270.045*m = E_k = \frac{1}{2}mv^2\\ 270.045 = \frac{1}{2}v^2 \mbox{ (divide both sides by m)}\\ v^2 = 2*270.045\\ v = \sqrt{2*270.045} \approx 23.2 m/s\\ \end{array}$$

So the answer to the first question is 23.2 m/s

Now for the second ball, let's pretend that the balcony is at height 0. That would mean that when the ball is thrown upwards with a speed of 13.1m/s it will reach a certain height. At this height, all kinetic energy will have transformed into potential energy.

If we want to calculate this height we can therefore again equate

$$E_k = E_p$$

$$\begin{array}{lcl} \frac{1}{2}mv^2 = mgh\\ \frac{1}{2}v^2 = gh\\ h = \frac{v^2}{2g}\\ h = \frac{13.1^2}{2*9.8}\\ h \approx 8.76\mbox{meters} \end{array}$$

Now let's get rid of the assumption that the balcony is at a height of 0.

We now know that the ball loses all it's kinetic energy at a height of 8.76+18.8 = 27.56 meters

This means at this height all it's energy has become potential energy.

We can calculate this potential energy with the equation

$$E_p = m*g*h = m*9.8*27.56 = 270.088*m$$

Again we know that when the ball is just a fraction from the ground all potential energy will have become kinetic energy.

Hence we can again equate this to

$$E_k = \frac{1}{2}mv^2$$

Solving this gives

$$\begin{array}{lcl} \frac{1}{2}{m}v^2 = 270.088*m\\ \frac{1}{2}v^2 = 270.088\\ v^2 = 2*270.088\\ v = \sqrt{2*270.088}\\ v \approx 23.2m/s\\ \end{array}$$

Since we have neglected air resistance here you can see that the law of thermodynamics apply.

Since both b***s have the same total energy when they leave the balcony, both b***s will have the same total energy when they hit the ground. Since in both cases this energy has totally become kinetic energy. the velocity of both b***s is the same when they hit the ground.

I'll have to look up some formula's with time before I'll be able to answer the other questions (since I haven't done this in a while).

Reinout

p.s. I might have made some round-off errors along the way since I've rounded off my answer several times during the calculations. My apologies for that.

Sep 16, 2014
#2
+2353
+5

All right,

so it's way more simpler than I anticipated it to be.

For the first ball we have the formula

$$v = at$$

where v is the change in speed, a is the acceleration and t is the time.

The change in speed for the first ball is

$$v = v_1-v_0$$

where v0 is the starting speed and v1 is the final speed.

therefore $$v= 23.2-13.1 = 10.1 m/s$$

a is equal to the gravitational force for which we used $$9.8m/s^2$$

Therefore $$t = \frac{v}{a} = \frac{10.1}{9.8} = 1.03\mbox{ seconds}$$

For the second ball there are two parts which we can calculate.

The part where the ball is thrown up and the speed changes from 13.1m/s to 0m/s

and the part where the ball goes down and the speed changes from 0m/s to 23.2 m/s

For the first part we have

$$v = ta$$

We know v = -13.1 and since the gravitational force is working against the speed we have a = -9.8

Therefore $$v = ta \Rightarrow t = \frac{v}{a} = \frac{-13.1}{-9.8} = 1.34 \mbox{ seconds}$$

For the second part we have

$$v = ta$$

v = 23.2m/s, and a = 9.8

Hence $$t = \frac{v}{a} = \frac{23.2}{9.8} = 2.37\mbox{ seconds}$$

Therefore the total time the second ball is in the air is given by

$$t = 1.34+2.37 = 3.71 \mbox{ seconds}$$

Reinout

Sep 16, 2014
#3
+27377
0

The useful formulae here are the constant acceleration kinematics formulae, namely:

1.  v = u + at

2.  v2 = u2 + 2as

3.  s = ut + (1/2)at2

u is initial velocity, v is final velocity, a is acceleration, t is time, s is distance.

So, we can use equation 2. to find the final velocity of the ball thrown vertically down (assuming no resistance due to air drag):

v2 = 13.12 + 2*9.8*18.8

$${\mathtt{v}} = {\sqrt{{{\mathtt{13.1}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{18.8}}}} \Rightarrow {\mathtt{v}} = {\mathtt{23.239\: \!836\: \!488\: \!237\: \!175\: \!2}}$$

So, velocity of first ball hitting ground is approximately 23.24m/s.

For the second ball the velocity as it hits the ground will be the same as it will lose speed as it rises and then regain the same speed by the time it gets back to the balcony, from where it will behave like the first ball to the ground!

To find the times use equation 3. for the first ball. and, for the second ball, equation 1. (setting v = 0) for the upward part of the flight and 3. for the downward part.

Sep 16, 2014
#4
+95360
0

Do you need it done by calculus?

Sep 16, 2014
#5
+2353
+5

For the last part, we again have

$$v = at$$

where v is the change in velocity.

For the ball going down, the change in velocity is given by.

$$v = 9.8*0.964 \approx 9.45 m/s$$

Since for the ball going upwards a = -9.8 and t = 0.964

the change in velocity for the ball going upwards is given by

$$v \approx -9.45m/s$$

For the ball going downwards, the speed has increased from 13.1 to 22.55m/s

This means that the current kinetic energy is given by

$$E_k = \frac{1}{2}mv^2 = \frac{1}{2}m(22.55)^2$$

Given question a, we knew the ball started with a total energy of

270.045 * m

Therefore the rest of the energy must be left as potential energy, or

$$\begin{array}{lcl} E_p = 270.045* m - \frac{(22.55)^2}{2}*m = mgh\\ 270.045-\frac{22.55^2}{2} = 9.8h\\ h = \frac{270.045-\frac{22.5^2}{2}}{9.8} \approx 1.73 \mbox{ meters}\\ \end{array}$$

For the ball going upwards we know the speed decreased from 13.1 to 13.1-9.45 = 3.65

The kinetic energy at a speed of 3.65m/s is given by

$$E_k = \frac{1}{2}mv^2 = \frac{m3.65^2}{2}$$

This means the rest has become potential energy.

Since we know $$E_t = 270.045$$

$$E_p = E_t-E_k = 270.045*m - \frac{m3.65^2}{2} = m*(9.8)*h$$

Therefore $$h = \frac{270.045-\frac{3.65^2}{2}}{9.8} \approx 26.88 \mbox{ meters}$$

The difference in height is therefore given by

26.88 - 1.73 = 25.15 meters

Reinout

Sep 16, 2014
#6
+27377
+5

Note that the kinematics equations arise very simply from calculus:

dv/dt = a

If a is constant then v = at + u (where u is v(0))

ds/dt = v so ds/dt = at + u so s = at2/2 +ut + s(0), and we usually set s(0) = 0, or redefine s as s-s(0), so

s = at2/2 + ut

We also have v2 = (at + u)= a2t2 + 2atu + u2 = 2a(at2/2 + ut) + u2 = 2as + u2

Sep 16, 2014
#7
+20850
+5

two students on a balcony 18.8m above the street. one student throws a ball vertically DOWNWARD 13.1 m/s; at the same instant, the onther student throws a ball vertically UPWARD at the same speed.

I. what is the volcity of the first ball as it strikes the ground? assume it is in the posotive direction... what is the volcity of the second ball as it strikes the ground?

$$g=9.81 \frac{m}{s^2} \quad v_0 = 13.1\frac{m}{s} \quad h=18.8\;m$$

$$\small{ \text{Way vertically downward: } s_1 = \frac{g}{2}t^2 \quad \text{and way thrown vertically downward or upward: } {s_2=v_0t\quad }$$

$$\small{ way \;first\; ball\; s_I:\boxed{s_I= s_1+s_2 = h} \quad way \;second\; ball s_{II}:\boxed{s_{II}= s_1-s_2 = h} }$$

$$\begin{array}{rcl|rcl} \small{ \text{time first ball:} \ t_1 } &&& \small{ \text{time second ball:} \ t_2 }\\ s_I= \frac{g}{2}t_1^2+v_0t_1 & = & h & s_{II}=\frac{g}{2}t_2^2-v_0t_2 &=&h\\ \frac{g}{2}t_1^2+v_0t_1 -h& = & 0 & \frac{g}{2}t_2^2-v_0t_2 -h&=& 0 \\ t_1>0: \quad t_1 &=& \frac{-v_0+\sqrt{v_0^2+2gh} }{g} &t_2>0: \quad t_2 &=& \frac{v_0+\sqrt{v_0^2+2gh} }{g} \\ \end{array}$$

$$\begin{array}{rcl|rcl} \small{ \text{final speed first ball:} \ v_1 } &&& \small{ \text{final speed second ball:} \ v_2 }\\ v_1&=& gt_1+v_0 & v_2 &=& gt_2-v_0 \\ v_1 & = & -v_0 + \sqrt{v_0^2+2gh} + v_0 & v_2 &=& v_0 + \sqrt{v_0^2+2gh} - v_0 \\ \end{array}$$

$$\boxed{v_1=v_2=\sqrt{v_0^2+2gh} } = \sqrt{13.1^2+2*9.81*18.8} = 23.2479246386\ \frac{m}{s}$$

II. what is the difference in the time the b***s spend in the air?

$$\begin{array}{rcl} \Delta t & = & t_2-t_1 \\ &=& \frac{(v_0 + \sqrt{v_0^2+2gh})}{g} - \frac{( -v_0 + \sqrt{v_0^2+2gh} ) } {g}\\ &=& \frac{2v_0}{g} \end{array}$$

$$\boxed{\Delta t=\frac{2v_0}{g} } = \frac{2*13.1}{9.81} = 2.67074413863\ s$$

III. how far apart are the b***s 0.964s after they are thrown?

$$\begin{array}{rcl} \Delta s & = & s_I-s_{II} \\ &=& (s_1 + s_2 ) - ( s_1 - s_2 )\\ &=&2s_2 \\ &=&2v_0t \end{array}$$

$$\boxed{ \Delta s=2v_0t } = 2*13.1*0.964 = 25.2568\ m$$

heureka Sep 17, 2014