A ball is thrown at a speed of 40 m/s at an angle of `60^0` with the horizontal. Find a. the maximum height reached and b. the range of te ball. Take `g=10 m/s^2`.

Answer

A, B, C, D

Explanation

Given `u=40 m/sa=g= 9.8 m/s^2, angle of projectiion
theta = 60^0`
a. Maximum height
`h= (u^2 sin^2 thetha)/(2g)
=(40^2(sin 60^0)^2)/(2xx10)= 60m
b. Horizontal range
x= (u^2 sin 2theta)/g
= (40^2 sin 2 xx 60^0)/g= 80 sqrt(3)m`