A ball is projected from a certain point on the surface of a planet at a certain angle with the horizontal surface. The horizontal and vertical displacements x and y vary with time t in second as `x=10sqrt(3)t` and `y=10t-t^(2)`
The maximum height attained by the ball is:

Options

A. 100m
B. 75m
C. 50m
D. 25m

Answer

D

Explanation

Here `v_(y)=d/(dt)(y)=d/(dt)(10t-t^(2))=10-2t`
At the highest point `v_(y)=0:.10-2t=0`
or `t=5s`
Putting above `h_(max)=10xx5-25=25m`