CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A body is dropped from the top of a tower It acquires a velocity $$20 ms^{-1}$$ on reaching the ground. Calculate the height of the tower (Take $$g=10ms^{-2}$$)


Solution

Given,
$$u=0m/s$$
$$v=20m/s$$
$$g=10m/s^2$$
From 3rd equation of motion,
$$2gh=v^2-u^2$$
$$h=\dfrac{v^2-u^2}{2g}$$
$$h=\dfrac{20\times 20-0\times 0}{2\times 10}$$
$$h=20m$$
The height of the tower is $$20m$$.

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image