Asked by Anonymous
In a tennis serve, a .074 kg ball can be accelerated from rest to 40 m/s over a distance of 0.79 m.Find the magnitude of the average force exerted by the racket on the ball during the serve.
Answers
Answered by
Henry
V ^2= 2ad =(40)^2,
2 * a * 0.79 = 1600,
1.58a = 1600,
a = 1600 / 1.58 = 1012.7 m/s^2.
F = ma = 0.74kg*1012.7 m/s^2 = 749.4N.
2 * a * 0.79 = 1600,
1.58a = 1600,
a = 1600 / 1.58 = 1012.7 m/s^2.
F = ma = 0.74kg*1012.7 m/s^2 = 749.4N.