Question

A 0.3 kg volleyball comes toward the volleyball net with a speed of 10 m/s at an angle of 12° below horizontal.  You jump up and hit the volleyball with your hand touching the ball for 0.1 seconds.  The volleyball then has a speed of 50 m/s and has a velocity direction that is 62° below horizontal.  What is the magnitude of the average force of your hand hitting the ball.

100 to 150 N

150 to 200 N

0 to 50 N

50 to 100

greater than 200 N

Answer 1

here,

the mass of volleyball , m = 0.3 kg

the initial velocity , u = 10 m/s * ( cos(12) i - sin(12) j)

u = (9.78 i m/s - 2.08 j m/s)

the final velocity , v = 50 m/s * ( cos(62) i - sin(62) j)

v = (23.47 i m/s - 44.15 j m/s)

time taken , t = 0.1 s

the average force applied , F = m * (v - u) /t

F = 0.3 * ( 23.47 i - 44.15 - 9.78 + 2.08) /0.1 N

F = (41.07 i - 126.21 j) N

the magnitude of the average force , |F| = sqrt(41.07^2 + 126.21^2) N

|F| = 132.7 N

the magnitude of the average force of your hand hitting the ball is a) (100 to 150 N)