Question

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 displacement X and Y vary with time t in second as : x=10√3 and y=10t-t^2 . Find the maximum height attained by the ball.

Answer 1

we have to find maximum height attain by ball .
so, we should use just y - component of motion of projectile.
given, y = 10t - t²
Here, y is function of time, ∴ for getting velocity we should differentiate y with respect to time .
e.g., dy/dt = 10 - 2t
at maximum height , velocity of ball = 0
so, dy/dt = Vy = 10 - 2t = 0 ⇒t = 5 sec

Now, maximum height attain by ball , y = 10t - t²
= 10(5) - (5)² = 50 - 25 = 25 m

Hence, maximum height = 25 m