Question

A cricket ball is thrown at a speed of 56m/s in a direction 30° above the horizontal. Calculate maximum height the time taken by the ball to return to the same level, and the horizontal distance from the thrower to the point where the ball returns to the same level.

Answer 1

If you take g=9.8ms^2 then we get the answer shown below, maximum height, H=u^2×sin^2X /2×g=56^2×sin30/9.8=40meter. Time taken,T= 2×u×sinx/g=2×56×sin30/9.8=5.71 second. Distance traveled, R=u^2×sin2x/g= 56^2×2sin30×cos30/9.8=277.1 meter...........is it correct answer. Please inform me........

Answer 2

Dear Student,

● Answer -

H = 40 m

R = 277 m

T = 5.714 s

◆ Explanation -

# Given -

u = 56 m/s

θ = 30°

# Solution -

Maximum height,

H = u²sin²θ/2g

H = 56² × sin²30 / 2×9.8

H = 40 m

Horizontal range,

R = u²sin2θ/g

R = 56² × sin(2×30) / 9.8

R = 277 m

Time of flight is -

T = 2usinθ/g

T = 2 × 56 × sin30 / 9.8

T = 5.714 s

Thanks dear. Hope this helps you...