The original question is- Julie throws a ball to her friend Sarah. The ball leav

Question

The original question is- 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 21 m/s at an angle 58 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

Then it changes and I am given- 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 14 m/s when it reaches a maximum height of 21 m above the ground.

So 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

distance between them two = u^2*sin(2theta)/g

X = 21^2*sin(2*58)/g

X = 40.45 mtr

t= X/Vx = 40.45/14 = 2.89s

y = h + Vy*t - 1/2 * gt^2

y = 1.5 + sqrt(2*9.8*19.5)*2.89 -4.9 *(2.89)^2

y = 17.07 m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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