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 17 m/s at an angle 36 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground. After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 17 m/s when it reaches a maximum height of 8 m above the ground. How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)

Explanation / Answer

1.

U = sqrt(Ux^2 + Uy^2); where Ux = 8 mps as the X component is constant. Uy = sqrt(2gh) to reach h = H - y = 11 - 1.5 meters above Sarah's hand. So U = sqrt(8^2 + 2*9.8*(11 - 1.5)) = 15.81771159 = 15.8 mps
so U=15.8 mps

2.
The distance between the two is X = U^2/g sin(2theta) = 14^2*sin(radians(2*62))/9.8 = 16.58075145 = 16.6 meters as the release and impact elevation are the same heights when Julie throws.

So Sarah's throw will be in the air t = X/Ux = 16.6/8 = ? seconds when it reaches Juilie's position. That will put the ball y(t) = h + Uy t - 1/2 g t^2 = 1.5 + sqrt(2*9.8*(11 - 1.5))*(16.6/8) - 4.9*(16.6/8)^2 = 8.716875072 = 8.7 m off the ground when it reaches Julie's ;position.

so the height of the ball= 8.7 m.

Related Questions

Navigate

• Browse (All)
• Browse J
• Subjects

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