Answer to Question #236671 in Physics for sam

Question

A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down, as shown in Figure. Using t_A= 0 as the time the stone leaves the thrower’s hand at position A, determine :

a. the time at which the stone reaches its maximum height,

b. the maximum height,

c. the time at which the stone returns to the height from which it was thrown,

d. the velocity of the stone at this instant, and

e. the velocity and position of the stone at t =5.00 s. 

Answer 1

Given:

$v_0=20.0\:\rm m/s$

$h=50.0\:\rm m$

a. the time at which the stone reaches its maximum height is given by

$$t=v_0/g=(20.0\:\rm m/s)/(9.8\:\rm m/s^2)=2.04\:\rm s$$

b. the maximum height is given by

$$h_{\max}=v_0^2/2g+h\\=(20.0\:\rm m/s)^2/(2*9.8\:\rm m/s^2)+50.0=70.4\:\rm m$$

c. the time at which the stone returns to the height from which it was thrown is given by

$$t_*=2t=2*2.04\:\rm s=4.04\: s$$

d. the velocity of the stone at this instant is equal to initial velocity

$$v=20.0\:\rm m/s$$

e. the velocity of the stone at t =5.00 s

$$v=v_0-gt=20.0-9.8*5=-29.0\:\rm m/s$$

the position of the stone at t =5.00 s

$$y=y_0+v_0t-gt^2/2$$

$$y=50+20*5-9.8*5^2/2=27.5\:\rm m$$