Hard
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 varies with time t in second as:
x = 10 and y = 10 √3t − t2
The maximum height attained by the ball is
x = 10 and y = 10 √3t − t2
The maximum height attained by the ball is
Options
A.100m
B.75 m
C.50 m
D.25 m
Solution
(Ymax) ⇒ 
⇒
(10 t − t2) = 10 − 2 t ⇒ t = 5
⇒ Ymax = 10(5) − 52 = 25 m Ans ″D″
⇒
(10 t − t2) = 10 − 2 t ⇒ t = 5⇒ Ymax = 10(5) − 52 = 25 m Ans ″D″
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