A ball is projected from a certain point on the surface of a planet at a certain angel with the horizontal surface. The horizontal and vertical displacement $x$ and $y$ vary with time $t$ in second as $x = 10\sqrt{3}t$ and $y = 10t - t^{2}$ The maximum height attained by the ball is

Question

Dr. Abubacker · Accepted Answer

(d) $v_{y} = \frac{d}{dt}(y) = \frac{d}{dt}(10t) - \frac{d}{dt}\left( t^{2} ight) = 10 - 2t$
At maximum height, $v_{y} = 0$
$	herefore\ 10 - 2t = 0$ or $2t = 10$ or $t = 5$ s
$	herefore\ y = (10 	imes 5 - 5 	imes 5)\ m = 25\ m$

Q. A ball is projected from a certain point on the surface of a planet at a certain angel with the horizontal surface. The horizontal and vertical displacement $x$ and $y$ vary with time $t$ in second as $x = 10\sqrt{3}t$ and $y = 10t - t^{2}$ The maximum height attained by the ball is

Explanation

Reviewed by Dr. Abubacker

Want more practice?