Aball rolls horizontally with a speed of 9.6 mis off of the edge of a tal platform. if the ball lands 5.7 m from the point on the ground directly below the edge of the platform, 1 what is the height of the platform? 2. how long is the ball in the air?

1) The height of the platform is 1.7 m

2) The ball is 0.59 s in the air.

Explanation:

The position of the ball at time "t" is given by the following position vector:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

r = vector position at time t

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² if the upward direction is considered positive)

Please, see the figure for a better understanding of the problem. Notice that the frame of reference is located at the edge of the platform. In this case, x0 = 0 and y0 = 0. Initially, the ball does not have a vertical velocity so that v0y = 0. The position vector will be as follows:

r = (v0x · t, 1/2 · g · t²)

Seeing the figure, notice that when the ball reaches the ground the x-component of its position vector will be 5.7 m. Then:

x = v0x · t

5.7 m = 9.6 m/s · t

t = 0.59 s

The ball reaches the ground in 0.59 s. At this time, the magnitude of the y-component of the position vector will be the height of the platform (see figure). Then:

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.59)²

y = -1.7 m

The height of the platform is 1.7 m.

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