CATAGORY: Mechanics QUESTION: A ball rolls off a cliff that is 10 m high. If the ball lands 5 m away from the edge a. how long does it take to hit the ground? b. what is it's initial speed? c. what will be it's final velocity (give both the magnitude and direction)

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Since the ball rolls of the cliff its initial vertical velocity will be horizontal. The vertical equation of motion will then be 1 (1) y = 10m - - g t^2 2 Where g = 9.8 m/s^2 is the magnitude of the acceleration due to gravity, t is time and we take zero altitude to be at the base of the cliff. The ball hits when y = 0 so 1 (2) 0 = 10m - - 9.8 m/s^2 ti^2 2 Where ti is the fall time. Solving for ti (3) ti = sqrt(2 x 10m / 9.8 m/s^2) = 1.429 s It takes 1.43 s to hit the ground. During this time it travels horizontally 5 m. The horizontal velocity (and also the initial velocity since this component does not change) is thus (4) horizontal velocity = 5.0 m / 1.429s = 3.499 m/s The vertical component of velocity ( since it has no initial vertical velocity) is given by (5) v vertical = - gt = -9.8m/s^2 (1.429s) = -14.004 m/s The magnitude of the final velocity is then v final = sqrt( (v horizontal)^2 + (v vertical)^2) = sqrt( (3.499 m/s)^2 + (14.004 m/s)^2) = sqrt( 208.355 (m/s)^2) = 14.435 m/s the direction that the ball impacts the ground is given by v horizontal ------> \ | \ | |v horizontal| \ | v vertical (6) tan(a) = -------------- \ | |v vertical| \a| \| v Where a is the angle FROM vertical. Thus (7) tan(a) = 3.499 / 14.004 = 0.2499 thus (8) a = 14.03 degrees To two significant figures then ************************************************* * (a) time of flight ~ 1.4 s * * * * (b) initial speed ~ 3.5 m/s (horizontal) * * * * (c) final velocty ~ 14 m/s at an angle * * of 14 degrees from the * * vertical. * * * *************************************************

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