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# phys

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A child throw a ball vertically upward from
the roof of a tall building at t=0.0 s with a velocity 25 m/s. The ball is the freely moves in the
air without friction. (a) Write down equations for the position y(t), instantenous velocity v(t)
& instantaneous acceleration a(t) , for time t. (b) Calculate the ball’s position, instantenous
velocity, & istantenous acceleration at time t for t=1s, 2s, 3s. (c) Find the time, t, when the
ball reaches mximum y position , y max . Also find y max . (d) Draw the path of the ball during
the time period, starting t=0 up to t=6s. Also interpret the motion of the ball. (e) Find the
direction, velocity ,& acceleration of the ball at t 6s.

Nov 27, 2020

#1
+10840
+1

A child throw a ball vertically upward from
the roof of a tall building at t=0.0 s with a velocity 25 m/s. The ball is the freely moves in the
air without friction. (a) Write down equations for the position y(t), instantenous velocity v(t)
& instantaneous acceleration a(t) , for time t. (b) Calculate the ball’s position, instantenous
velocity, & istantenous acceleration at time t for t=1s, 2s, 3s. (c) Find the time, t, when the
ball reaches mximum y position , y max . Also find y max . (d) Draw the path of the ball during
the time period, starting t=0 up to t=6s. Also interpret the motion of the ball. (e) Find the
direction, velocity ,& acceleration of the ball at t 6s

Hello Guest!

$$(a)\\ y(t)=v\cdot t-\frac{1}{2}gt^2\\ v=v_0-g\cdot t\\ a=g=9.81\frac{m}{s^2}\ (constant)$$

$$(b)\\ y(t)=v\cdot t-\frac{1}{2}gt^2\\y(1s)=25\frac{m}{s}\cdot1s- \frac{1}{2}\cdot9.81\frac{m}{s^2}\cdot 1^2s^2=20.095m\\ y(2s)=25\frac{m}{s}\cdot2s- \frac{1}{2}\cdot9.81\frac{m}{s^2}\cdot 2^2s^2=30.38m\\ y(3s)=25\frac{m}{s}\cdot3s- \frac{1}{2}\cdot9.81\frac{m}{s^2}\cdot 3^2s^2=30.855m$$

$$v=v_0-gt\\ v(1s)=25\frac{m}{s}-\frac{9.81m\cdot 1s}{s^2}=15.19\frac{m}{s}\\ v(2s)=25\frac{m}{s}-\frac{9.81m\cdot 2s}{s^2}=5.38\frac{m}{s}\\ \color{blue}v(2.548s)=25\frac{m}{s}-\frac{9.81m\cdot 2.548s}{s^2}=0\frac{m}{s}\\ \color{blue}v(5.097s)=25\frac{m}{s}-\frac{9.81m\cdot 5.097s}{s^2}=-25\frac{m}{s}\\ v(6s)=25\frac{m}{s}-\frac{9.81m\cdot 6s}{s^2}=-33.86\frac{m}{s}$$

$$s=v\cdot t-\frac{1}{2}gt^2=0$$

$$v\cdot t=\frac{1}{2}gt^2\\ \frac{1}{2}gt^2-v\cdot t=0\\ \color{blue}t_1=0\\ \frac{1}{2}gt=v\\ t_2=\frac{2v}{g}=\frac{2\cdot 25m\cdot s^2}{s\cdot 9.81m}$$

$$t_2=5,097s$$

$$t_{max}=\frac{t_2}{2}=\frac{5.097s}{2}$$

$$t_{max}=2.548s$$

$${\color{blue}y_{max}}=25\frac{m}{s}\cdot2.548s- \frac{1}{2}\cdot9.81\frac{m}{s^2}\cdot 2.548^2s^2=\color{blue}31.855m$$

$$y_{max}=31.855m$$

$$\color{blue} y(5.097s)=25\frac{m}{s}\cdot5.097s- \frac{1}{2}\cdot9.81\frac{m}{s^2}\cdot 5.097^2s^2=0\ m\\ y(6s)=25\frac{m}{s}\cdot6s- \frac{1}{2}\cdot9.81\frac{m}{s^2}\cdot 6^2s^2=-26.58m$$

The acceleration at any point in time and at any position of the ball on its path is

$$g=9.81m/s^2$$.

!

Nov 27, 2020
edited by asinus  Nov 27, 2020