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 10m/s at an angle 37 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground. What is the horizontal component of the ball's velocity when it leaves Julie's hand? What is the vertical component of the ball's velocity when it leaves Julie's hand? What is the maximum heights the ball goes above the ground? What is the distance between the two girls? 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 11 m/s when it reaches a maximum height of 5 m above the ground. What is the speed of the ball when it leaves Sarah's hand? 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

part 1 )

horizontal component = vx = v*costheta

vx = 10*cos37 = 7.986 m/s

part 2 )

vertical component = vy = v*sintheta

vy = 10*sin37 = 6.018 m/s

part 3 )

h = vy^2/2g

h = 1.85 m

part 4 )

R = v^2(sin2theta)/g

R = 9.8 m

part 5 )

v = sqrt(vx^2 + vy^2)

vx = 7.986 m/s

vy = sqrt(2gh)

here h = 5-1.5 = 3.5

v = 11.5 m/s

part 6 )

t = R/vx = 1.227 s

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

y = 1.5 + sqrt(2*g*3.5) * 1.227 - 1/2 * 9.8 * 1.227^2

y = 4.2855 m

Related Questions

Navigate

• Browse (All)
• Browse J
• Subjects

---

---

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