Answer:
maximum horizontal distance = 10m
initial vertical velocity of the ball = 4.9m/s
Explanation:
Complete question
<em>A ball is launched with an initial horizontal velocity of 10.0 meters per second. It takes 500 milliseconds for the ball to reach its maximum height.
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<em>Determine the maximum horizontal distance that the ball will travel.
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<em>Calculate the initial vertical velocity of the ball.</em>
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Maximum horizontal distance x is expressed as;
x = vT
T is total time of flight
T = 2t
Hence x = 2vt
v is the velocity
t is the time
Given
v = 10.0m/s
time t = 500ms = 0.5s
Horizontal distance = 2 * 10 * 0.5
Horizontal distance = 20 * 0.5
Horizontal distance = 10m
Hence the maximum horizontal distance that the ball will travel is 10m
To get the initial horizontal distance, we will use the equation of motion
v = u - gt
T maximum height, v = 0
Substitute
0 = u - 9.8(0.5)
-u = - 4.9
u = 4.9m/s
Hence the initial vertical velocity of the ball is 4.9m/s
