Question

5. A batsman hits a ball at an angle of 30° with an
initials speed of 30 ms". Assuming that the ball
travels in a vertical plane, calculate
i) The time at which the ball reaches the
highest point
ii) The maximum height reached
iii) The horizontal range of the ball
iv) The time for which the ball is in the air​

Answer 1

Explanation:

Here θ=30∘,u=30m/s

(a) The time taken for ball to reach 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  

tH=T2=u sinθ/g=30/10×sin30∘=1.5s  

(b) The maximum height  is  

u^2sin^2θ / 2g=(30)^2×(sin30∘)^2 /2g=9002×10×4=11.25m  

( c) Horizontal range =u^2sin2θ / g  

=(30)^2 sin2(30∘) / 10  

=900√3 / 20m =45√3m  

(d) The time for which the ball is in air is same as its time of flight, i.,e.,  

2usinθ / g=2×30×sin30∘ / 10=3s.

THANKS HOPE IT HELPS U

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