Question

A ball is thrown horizontally from the top of a tower with a velocity of 40m/s take g=10m/s if the ball reaches the ground in 4second find height of the tower ??

Answer 1

U= 0m/s
A= 10m/s
T= 5s
V= 40m/s
S= ut + 1/2at^2
= 0*5+ 1/2*10*4^2
= 0+ 10*8
= 80m
So height is 80m

Answer 2

The height of the tower is $80m$.

Step by step calculation:

Given data:

Velocity of the ball$=40m/s$

g$=10m/s$

time taken for the ball to reach ground$=4s$

To find:

We must find the height of the tower.

Formula to be used:

$s=ut+(1/2)at^{2}$

s$=height$

$u=$initial velocity

$t=time$

a$=acceleration$

Solution:

Using the formula $s=ut+(1/2)at^{2}$

$a=-10m/s^{2}$ since acting downwards.

$s=40$×$4-$$(1/2)$×$10$×$4^{2}$

$s=160-80$

$s=80m$

The height of the tower is $80m$

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