(a) The maximum height of the ball above the ground is 12.5 m and the time of motion is 1.43 s.
(b) The time taken for the ball to contact the other player at 0.5 m above the ground is 3.0 s.
The maximum height reached by the ball is calculated as follows.
At maximum height, the final velocity, v = 0
dh/dt = v = 0
dh/dt = -2(4.9)t + 14
0 = -9.8t + 14
9.8t = 14
t = 1.43 s
H(1.43) = -4.9(1.43)² + 14(1.43) + 2.5
H(1.43) = -10.02 + 20.02 + 2.5
H(1.43) = 12.5 m
12.5 = -4.9t² + 14t + 2.5
4.9t² - 14t + 10 = 0
t = 1.43 s
0.5 = -4.9t² + 14t + 2.5
4.9t² - 14t + - 2 = 0
t = 3.0 seconds
The complete question is below:
In a volleyball match Hanein serves the volleyball at 14 m/s, from a height of 2.5 m above the court. The height of the ball in flight is estimated using the equation, h = -4.9t² + 14t + 2.5, where t is the time in second and h is height above ground, in metres.
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