Question

a ball is thrown vertically upwards from the top of a tower with initial velocity 19.6m/s. It reaches the ground after 5 seconds. Calculate

i.the height of the tower

ii.the velocity of the ball on reaching the ground (take g = 9.8m/s^2)

Answer 1

Answer:

(I). The height of the tower is 18.38 m.

(II). The velocity of the ball on reaching the ground is 4.88 m/s.

Explanation:

Given that,

Initial velocity = 19.6 m/s

Time t = 5 sec

(I). We need to calculate the height

Using equation of motion

$h = ut+\dfrac{1}{2}gt^2$  

Where, h = height

t = time

Put the value into the formula

$h =19.6\times2.5-\dfrac{1}{2}\times9.8\times(2.5)^2$

$h=18.38\ m$

(II). We need to calculate the velocity of the ball on reaching the ground

Using equation of motion

$v^2=u^2+2gh$

Put the value into the formula

$v^2=(19.6)^2-2\times9.8\times18.38$

$v^2=23.912\ m/s$

$v=4.88\ m/s$

Hence, (I). The height of the tower is 18.38 m.

(II). The velocity of the ball on reaching the ground is 4.88 m/s.