Question

A cricketer hits a ball with a velocity 20 m/s at 60" above the horizontal. Find the Maximum height attained by the ball (in m). (g = 10 m/s)​

Answer 1

Answer:

Explanation:8.2 m

Equation of trajectory when a projectile is thrown with velocity u at an angle θ with the horizontal is

y=xtanθ−

2

1

​

u

2

cos

2

θ

gx

2

​

y=50tan60

0

−

2

1

​

×

25×25×[

2

1

​

]

2

9.8×50×50

​

=50

3

​

=9.8×8

=86.60−78.4=8.2m