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# Projectile motion

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In this question, use g = 10m*s^-2.

In cricket, a fast bowler projects a ball at 40m*s^-1 from a point h m above the ground, which is horizontal, and at an angle alpha above the horizontal. The trajectory is such that the ball will strike the stumps at ground level a horizontal distance of 20 m from the point of projection.
(i) Determine, in terms of h, two possible values of tan alpha. Explain which of these two values is the more appropriate one, and deduce that the ball hits the stumps after approximately half a second.
(ii) State the range of values of h for which the bowler projects the ball below the horizontal.
(iii) In the case h - 2.5, give an approximate value in degrees, correct to two significant figures, for alpha. You need not justify the accuracy of your approximation.

https://brainmass.com/physics/resistance/physics-projectile-motion-problem-188824

#### Solution Preview

i) Initial velocity can be resolved into x and y components :

ux = 40cosα
uy = 40sinα

Let the ball strike the base of the stumps after t secs. Vertical displacement when the ball strikes the base of the stumps = - h

Applying s = ut + ½ at2 for the vertical motion we get :

-h = 40sinα T - ½ g T2 = 40sinα T - ½ x 10 x T2 = 40sinα T - 5 T2 .......(1)
where T is the time of flight.

Or sinα = (5T2 - h)/40T ...

#### Solution Summary

This solution helps with a problem on projectile motion of a cricket ball. A step by step solution provided.

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