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 seconds) as x=10√3t and y=10t-t2 . The maximum height attained by the ball is

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 seconds) as x=10√3t and y=10t-t2 . The maximum height attained by the ball is (A) 100m (B) 75 m (C) 50 m (D) 25 m

Tardigrade Admin · Accepted Answer

( textdy/ textdt) = textv texty = text10 - text2t at maximum height textvy = 0 t = 5 s So maximum height attained ⇒ y textat = 5 = text10 t - textt2 = 50 - 25 = 25 m

Q. 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 seconds) as $x = 103 t$ and $y = 10 t - t^{2}$ . The maximum height attained by the ball is

$100 m$

$75 m$

$50 m$

$25 m$

$\frac{dy}{dt} = v_{y} = 10 - 2t$
at maximum height $v_{y} = 0$
$t = 5 s$
So maximum height attained $\Rightarrow y_{at = 5} = 10 t - t^{2}$
$= 50 - 25$
$= 25 m$

Q. 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 seconds) as x=103​t and y=10t−t2 . The maximum height attained by the ball is

100m

75m

50m

25m