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You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at the point even with the roof railing with an upward speed of 15.0 m/s, the ball is then in free fall. On its way back down, it just misses the railing. At the location of the building, g = 9.8m/s^2.

 

Find :

(a) the position and the velocity of the ball 1.0s and 4.0s after leaving your hand.

(b) The velocity when the ball is 5.0m above the railing.

(c) The maximum height reached and the time at which it is reached. 

(d) The acceleration of the ball when it is at its maximum height.

(e) Find the time when the ball is 5.0m below the roof railing.

Guest Jul 31, 2017
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 #1
avatar+7059 
+1

Hello Guest!

 

You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at the point even with the roof railing with an upward speed of 15.0 m/s, the ball is then in free fall. On its way back down, it just misses the railing. At the location of the building, g = 9.8m/s^2.

 

Find :

(a) the position and the velocity of the ball 1.0s and 4.0s after leaving your hand.

(b) The velocity when the ball is 5.0m above the railing.

(c) The maximum height reached and the time at which it is reached. 

(d) The acceleration of the ball when it is at its maximum height.

(e) Find the time when the ball is 5.0m below the roof railing.

 

(a) the position and the velocity of the ball 1.0s and 4.0s after leaving your hand.

 

\(h=vt-\frac{1}{2}gt^2\\ h_{1.0}=\frac{15m}{s}\cdot 1s-\frac{1}{2}\cdot\frac{9.8m}{s^2}\cdot 1^2s^2\\ \color{blue}h_{1.0}=10.1m\ over\ the\ railing\)

 

\(h_{4.0}=\frac{15m}{s}\cdot 4s-\frac{1}{2}\cdot\frac{9.8m}{s^2}\cdot 4^2s^2\\ h_{4.0}=60m-79.2m=-19.2m\\ \color{blue}h_{4.0}=19.2m\ under\ the\ railing\)

 

\(v=v_0-gt\\ v_{1.0}=\frac{15m}{s}-\frac{9.8m\cdot1s}{s^2}=5.2\frac{m}{s}\\ \color{blue}v_{1.0}=5.2\frac{m}{s}\ up \)

 

\(v_{4.0}=\frac{15m}{s}-\frac{9.8m\cdot4s}{s^2}=-24.2\frac{m}{s}\\ \color{blue}v_{4.0}=24.2\frac{m}{s}\ down\)

 

(b) The velocity when the ball is 5.0m above the railing.

 

\(v=v_0-\sqrt{2gh}\\ v_{(+5m)}=\frac{15m}{s}-\sqrt{\frac{2\cdot 9.8m\cdot 5m}{s^2}}=5.1\frac{m}{s}\\ \color{blue}v_{(+5m)}=5.1\frac{m}{s}\ up\)

 

 

(c) The maximum height reached and the time at which it is reached

 

\(E=\frac{mv^2}{2}=mgh\\ h=\frac{mv^2}{2mg}=\frac{v^2}{2g}\)

 

\(h_{max}=\frac{15^2m^2s^2}{2\cdot 9.8ms^2}\\ \color{blue} h_{max}=11.480m\ over\ the\ railing. \)

 

\(v=gt\\ t_{rise}=\frac{v}{g}=\frac{15ms^2}{9.8ms}\\ \color{blue} t_{rise}=1.531s\)

 

(d) The acceleration of the ball when it is at its maximum height.

 

The ball is weightless throughout the journey.
For the ball, the acceleration is zero everywhere. It is not exposed to any force.
For the viewer the acceleration is at the highest point and in the case g = 9.8m / s²

 

(e) Find the time when the ball is 5.0m below the roof railing.

 

\(s=vt-\frac{1}{2}gt^2\\ \frac{1}{2}gt^2-vt+s=0\\ t^2-\frac{2v}{g}t+\frac{2s}{g}=0\)

 \(t=-\frac{p}{2}+\sqrt{(\frac{p}{2})^2-q}\\ \color{blue} t=\frac{v}{g}+\sqrt{\frac{v^2}{g^2}-\frac{2s}{g}}\)    

\(t=\frac{15m/s}{9.8m/s^2}+\sqrt{(\frac{15}{9.8}s)^2-\frac{-2\cdot 5m}{9.8m/s^2}}\\ t=1.531s+\sqrt{2.343s^2+1,020s^2}\\ t=1.531s+1.834s\\ \color{blue}t=3.365s\ (5m\ under\ the\ railing)\)

 

laugh  !

asinus  Jul 31, 2017
edited by asinus  Jul 31, 2017
edited by asinus  Aug 1, 2017
edited by asinus  Aug 1, 2017
 #2
avatar+90565 
0

Thanks asinus :)

 

Guest:

Are you supposed to use physics formulas or calculus ?

Do you understand asinus's answer and is that what you were after ?

Melody  Jul 31, 2017
 #3
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0

Physics.

Guest Aug 1, 2017
 #4
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0

Deleted. Malfunction in presentation. 

Guest Aug 1, 2017
edited by GingerAle  Aug 1, 2017
 #5
avatar+725 
+1

Solutions for A - E

 

\(\text{Position and Time }\\ \begin{array}{|rcll|} \hline a_1) \text { position }\\ h &=& v_i(t) -\dfrac{1}{2} g(t)^2 \\ h &=& 15t - 4.9t^2 \\ t\tiny \; \text{ @} \small \text { 1 and 4}\\ h &=& 15(1) - 4.9(1)^2 &=&10.1m \\ h &=& 15(4) - 4.9(4)^2 &=&-18.4m \\ a_2) \text { velocity} \\ v_f &=& v_i – gt \\ v_f &=& 15 - 9.8t \\ \text{t @ 1 and 4} \\ v_f &=& 15 - 9.8(1) &=&5.20m/s \\ v_f &=& 15 - 9.8(4) &=& -24.2m/s \\ \hline \end{array}\\ \text { at 1 second: ball is 10.1m (above) and moving @ 5.20 m/s (up)} \\ \text { at 4 seconds: ball is -18.4m (below) and moving -24.2 m/s(down)} \\\)

 

\(b)\text {Gravity is slowing ball} \\ \begin{array}{|rcll|} \hline v^2 &=&v_0^2 +2(a)(\Delta y) \\ v^2 &=&(15)^2 +2(-9.8)(5m) \leftarrow \text{ use negative acceleration} \\ v &=&\pm \sqrt{127} \text{ (use positive result) } \\ v &=& 11.27m/s \\ \hline \end{array}\)

 

\(c)\text {Maximum height occurs when the vertical velocity is zero (0)} \\ \begin{array}{|rcll|} \hline a &=& \dfrac {(v_f – v_i)}{t} \\ t &=& \dfrac {(v_f – v_i)}{a} \\ t &=& \dfrac {(0 - 15)}{(- 9.8)} \\ t &=& \dfrac{-15}{-9.8} \\ t &=& 1.53s \\ \hline \end{array}\)

 

\(d) \text{ Acceleration is constant (g)} \\\)

 

\(e) \text{ Time after displacement }\\ \begin{array}{|rcll|} \hline t &=& \dfrac {(v_f – v_i)}{a} \\ v_f^2 &=& v_i^2 + 2(a)(\Delta y)\\ v_f &=& \pm \ sqrt{(v_i^2 + 2(a)(\Delta y)} \leftarrow \small \text {Use negative result; the ball is moving downward } \\ t &=& \dfrac{-\sqrt{(v_i^2 + 2(a)(\Delta y)} + (v_i)}{(-g)} \\ t &=& \dfrac{-\sqrt{(15^2 + 19.6*5)} + 15)}{-9.8}\\ t &=& 3.36s \\ \hline \end{array}\)

 

 

For related effects of gravity, see this.

GingerAle  Aug 1, 2017

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