Question

A ball is thrown with an initial speed at an angle with the horizontal the horizontal range of the ball is R and the ball reaches a maximum height R/6. In terms of R and g find the time interval during which the ball is in motion?

Answer 1

#sqrt((4R)/(3g))

Explanation:

Let the speed of projection be `u` and the angle of projection be `alpha`.

Then,

• the range is `R = u^2/g sin(2alpha)`
• the maximum height is `H = u^2/(2g)sin^2alpha`
• the time of flight is `T = (2u)/g sin alpha`

Thus

`T^2 = 4u^2/(g^2) sin^2 alpha = 8/g u^2/(2g)sin^2 alpha`

and hence

`T^2 = 8H/g`

Since `H` is given by `R/6`, we have

`T^2 = 8/g R/6 = (4R)/(3g)`