Question

a batsman hits a ball with a velocity of 28 ms-1 at an angle of 30° from the horizontal. If the boundary line is 60 m away in the direction where the batsman hit the ball, will the ball cross the boundary line?
[Use R = Vo^2 sin200/g]​

Answer 1

Explanation:

Here θ=30o,u=30ms−1

a. The time taken by the ball to reach the highest point is half the total time of flight. As the time of ascending and descending is same for a projectile without air resistance. the time to reach the highest point T

tH=2T=gusinθ=1030×sin30o=1.5s

b. The maximum height reached is

2gu2sin2θ=2g(302)×(sin30o)2=2×10×4900=11.25m

c. Horizontal range = gu2sin2θ

= 10(30)2sin2(30o)

= 209003=453m

d. The time for which thc ball is in air is same as its time of eight, i.e.,

g2usinθ=102×30×sin30o=3s

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