Question

A ball is thrown vertically upwards from the top of a tower with a speed of 100m/s.it strikes the pound near the base of the tower after 25sec . The height of the tower is

Answer 1

Answer:

562.5 m

Explanation:

The vertical position of the ball is given by

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

where

h is the height of the tower (the initial position of the ball)

u = 100 m/s is the initial velocity of the ball (positive because it points upwards)

g = -9.8 m/s^2 is the acceleration due to gravity (negative because it points downward)

t is the time

Substituting t =25 s and y = 0 (at t=25 s, the height of the ball is zero, since it has reached the base of the tower), we can re-arrange the equation to find h:

$h=y-ut-\frac{1}{2}gt^2 = 0-(100 m/s)(25 s) - \frac{1}{2}(-9.8 m/s^2)(25 s)^2 = 562.5 m$