A gun is fired straight into the air, and the height to which the bullet rises i

Question

A gun is fired straight into the air, and the height to which the bullet rises is measured. The gun is then rotated an angle theta down from the vertical and is fired again. This time the horizontal distance that the bullet travels as it rises and returns to its original height is measured, and it comes out to the same value as the height measured previously. Assuming the gun fires with the same muzzle velocity each time, and ignoring air resistance, which of the following values of theta will produce this result?a.15degtee b.30degtee c.45degtee d.60degtee e.none of these

Explanation / Answer

when fired straight in to the air

let initial velocity is v

height measured is h = v^2/(2g)

when fired at angle (90 - theta)degree with the horizontal with same velocity v

horizontal velocity Vh = vcos(90 - theta) = vsin(theta)

vertyical velocity Vv = vsin(90 - theta) = vcos(theta)

time for entire motion = t = 2*(Vv)^2/(2g) = [vcos(theta)]^2/g

so horizontal distance covered = D = t*Vh = [vcos(theta)]^2*vsin(theta)/g = (v^3)cos^2(theta)*sin(theta)/g

and given is h = D

=> v^2/2g = (v^3)cos^2(theta)*sin(theta)/g

=> 0.5 = v*cos^2(theta)sin(theta)

HERE WE REQUIRE INITIAL VELOCITY TO FIND OUT THETA.

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