10 A projectile is fired in such a way that its horizontal range is equal to thr

Question

10 A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection? Give your answer to three sig nificant figures. A ball is tossed from an upperstory window of a build. ing. The ball is Riven an initial velocity of 8.00 m/satan angle of 20.0" below the horizontal. It strikes the ground 3.00 Slater (a) How far horizontally from the base of the building does the ball strike the ground? (b) Find the height from which the ball was thrown. (c) How long does it take the ball to reach a point 10.0 m below the level of launching? 14, A 175 kg ball on the end of a string is revolving uniformly in a horizontal circle of radius 0.500 m. The ball makes 2.00 revolutions in a second. (a) Determine the speed of the ball. (b) Determine the ball's centripetal acceleration.

Explanation / Answer

10) Range = u^2 sin(2 theta) / g

maximum height = u^2 sin^2(theta) / 2g

Given that

Range = 3 x Max Height

u^2 sin(2 theta) / g = 3 x u^2 sin^2(theta) / 2g

4 cos(theta) = 3 sin(theta)

tan(theta) = 4/3

theta = arctan(4/3) = 53.1 degrees

Angle of projection = 53.1 degrees

13) u = 8 m/s

theta = 20 degrees

Horizontal distance travlled = u cos(theta) t = 8 * cos(20 degrees) * 3 = 22.55 m

H = u sin(theta) t + 0.5 g t^2

H = 8 * sin(20 degrees) * 3 + ( 0.5 * 9.81 * 3 * 3 ) = 52.35 m

Height = 52.35 m

10 = 8 *sin(20 degrees) * t + 0.5 * 9.81 * t^2

Solving above equation we get t = 1.17 s

Time taken = 1.17 s

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14) speed of ball = v = w r = ( 2 * pi ) * 0.5 = 3.14 m/s

centripetal acceleration = v^2 / r = 3.14^2 / 0.5 = 19.74 m/s2

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