Question

A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20 m above the ground. Determine the following: (a) The velocity v and elevation y of the ball above the ground at any time t. (b) The highest elevation reached by the ball and its corresponding time t. (c) The time when the ball will hit the ground and the impact velocity.

Answer 1

Answer:

Explanation:

Given

Initial velocity of ball $u=10\ m/s$

height of window $h=20\ m$

Using Equation of motion

$y=ut+\frac{1}{2}at^2$

where u=initial velocity

t=time

a=acceleration

As ball is already is at a height of 20 m so

$Y=ut+\frac{1}{2}at^2+20$

$Y=10\times t+0.5\times (-9.8)t^2+20$

$Y=-4.9t^2+10t+20$

(b)highest point is obtained at v=0

$v^2-u^2=2as$

where

v=final velocity

u=initial velocity

a=acceleration

s=displacement

$(0)-10^2=2\times (-9.8)\times s$

$s=\frac{100}{19.6}$

$s=5.102\ m$

Highest Point will be $s+20=25.102\ m$

(c)Time taken when the ball hit the ground i.e. at Y=0

$-4.9t^2+10t+20=0$

$t=3.28\ s$

impact velocity $v=\sqrt{2\times 9.8\times 25.102}$

$v=22.181\ m/s$