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 20 m/s at an angle 51 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?2)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 16 m above the ground, and takes 2.659 s to get directly over Julie's head.What is the speed of the ball when it leaves Sarah's hand?3)How high above the ground will the ball be when it gets to Julie?

Explanation / Answer

(a) Since the heights are same, by symmetry, the velocity when Sarah catches the ball would be 20 m/s at an angle of (-51) degrees from the horizontal. In other words, the vertical component of the velocity will be reversed.

So the vertical component of velocity is (20 * sin(51degrees)) in downward direction.

ans: 15.543 in downward direction.

(b) In the first case, the time of flight is (2*v*sin(51))/g (this is by calculating the change in vertical velocity and dividing by accelaration). So the distance between Sarah and Julie is (v*cos(51)) multiplied by time i.e, (2 * v*cos(51)*v*sin(51))/g i.e, (v*v*sin(102))/g i.e, 39.92 metres.

The horizontal velocity when Sarah throws the ball is (39.92/2.659) m/s i.e, 15.013 m/s towards Julie.

The vertical velocity when Sarah throws the ball is (by using second equation of motion), (h+0.5*g*t*t)/t, h would be 16 - 1.5 = 14.5. So the vertical velocity is 18.482 m/s up

So the speed would be 23.81 m/s.

(c)

y = yo + sqrt(2gh)*t - 1/2*g*t^2

y = 1.5 + sqrt(2*9.8*14.5)*2.659 - 4.9 * 2.659^2

y = 11.68 m

Related Questions

Navigate

• Browse (All)
• Browse J
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.