Question

If T be the total time of flight of a current of water and H be the maximum height attained by it from the point of projection, then HIT will be : (u = projection velocity, e = projection angle)

Options

• `(1/2)`u sin θ

• `(1/4)`u sin θ

• u sin θ

• 2u sin θ

Answer 1

`bb((1/4)u  sin  θ)`

Explanation:

Given: T – time period, H – maximum height

`"H"/"T" = (("u"^2sin^2theta)/(2"g"))/((2"u"sintheta)/g)`

= `1/4` u sin θ.