Question

a body is projected from the top of the tower 50m high with a velocity of 10m/s. calculate after how long and with what velocity will the body hit the ground. at what distance away from the foot of the tower will the body hit the ground

Answer 1

given:-

height = 50m

initial velocity = 10m/s

to find:-

time of flight

range of projection.

solution:-

use the equation:

$\boxed{\red{\tt S=ut+1/2at^2}}$

$\tt\implies -50=0+1/2(-10)t^2$

$\tt\implies\cancel{ -}50=\cancel{-}5t^2$

$\tt \implies 10=t^2$

$\boxed{\tt time=✓10sec }$

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for finding final velocity,

ux = 10m/s

uy= 0m/s

apply the formula:

$\boxed{\pink{\sf v^2-u^2=2as}}$

velocity in the x direction remains constant.

$\tt v^2=2(10)(50)$

$\tt v=✓1000$

$\tt v=10✓10 m/s$

range = ux×t

range = 10×√10

range = 10√10 m

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hope this helps!!