Asked by Jessica
A volleyball is launched straight up at an initial velocity of 2.3 m/s. It is then spiked by a taller player 0.8 meters above the launch height. How high did it rise, and how long did the volleyball remain in the air between the players?
Answers
Answered by
Henry
V^2 = Vo^2 + 2g*h = 0 @ max ht.
h max = -(Vo^2)/2g = -(2.3^2)/-19.6 = 0.270 m.
Since the max ht. is only 0.27 m, the
ball cannot be spiked at o.8 m. Please
check the numbers for accuracy.
h max = -(Vo^2)/2g = -(2.3^2)/-19.6 = 0.270 m.
Since the max ht. is only 0.27 m, the
ball cannot be spiked at o.8 m. Please
check the numbers for accuracy.
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