Part 1: A ball is thrown horizontally from the top of a building 28.6 m high. Th

Question

Part 1:
A ball is thrown horizontally from the top of a building 28.6 m high. The ball strikes the ground at a point 80.3 m from the base of the building.

The acceleration of gravity is 9.8 m/s^2. Find the time the ball is in motion. Answer in units of s.

Part 2:
Find the initial velocity of the ball. Answer in units of m/s.

Part 3:
Find the x componet of its velocity just before it strikes the ground. Answer in units of m/s.

Part 4:
Find the y componet of its velocity just before it strikes the ground. Answer in units of m/s.

Explanation / Answer

Height h = 28.6 m Range R = 80.3 m (1). Time of the ball in motion t = sqrt [ 2h / g ] = 2.415 s (2). Initial velocity of the ball v = R / t = 33.23 m / s (3). X-component of velocity when it hits the ground u = v = 33.23 m / s Since there is no acceleration along x- direction. (4). Y-component of velocity when it hits the ground V = gt = 23.667 m / s Since initial velocity of the along vertical direction is zero

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