Answer:
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
<u>t = 0.59 s</u>
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.
