Julie throws a ball to her friend Sarah. The ball leaves Julie\'s hand a distanc

Question

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 16 m/s at an angle 28 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

1)

What is the vertical component of the ball’s velocity right before Sarah catches it?

m/s

2) What is the time the ball is in the air?

s

3) What is the distance between the two girls?

m

4) After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 7 m above the ground, and takes 1.352 s to get directly over Julie's head.

What is the speed of the ball when it leaves Sarah's hand?

Explanation / Answer

a) Vy= 16 sin 28 = 7.511 m/s apprx

b) Time ball is in the air= ( Vinal - Vinitial)/g = ( -7.511- 7.511)/9.8 = 1.5329 sec apprx

c)Distance etween two girls = range= total time ( horizontal componnet of velocity ) = 1.5329( 16 cos 28) = 21.6555 m appex

d) vertical distanmce of ball above Julie's head= 7 - 1.5 = 5.5 m

ball's velocity when it's above jUlie head = sqr00t( 2gh) = sqroot( 2 x 9.8 x 5.5) = 10.382 m/s apprx

horizontal component of velocity when it reaches Julie's head = range/ time= 21.6555/ 1,.352= 16.02 m/s apprx

Velocity = sqroot( 16.02^2 + 10.382^2) = 19.087 m/s apprx

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