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Answer:
maximum horizontal distance = 10m
initial vertical velocity of the ball = 4.9m/s
Explanation:
Complete question
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.
Determine the maximum horizontal distance that the ball will travel.
Calculate the initial vertical velocity of the ball.
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