Posted by **aaa** on Saturday, September 17, 2011 at 1:17pm.

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 12 m/s at an angle 35 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

1)What is the horizontal component of the ball’s velocity when it leaves Julie's hand?

2)What is the vertical component of the ball’s velocity when it leaves Julie's hand?

3)What is the maximum height the ball goes above the ground?

4)What is the distance between the two girls?

5)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 13 m/s when it reaches a maximum height of 7 m above the ground.

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

6)How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)?

- physics -
**Henry**, Monday, September 19, 2011 at 5:04pm
Vo = 12m/s @ 35 deg.

1. Xo = hor. = 12cos35 = 9.83m/s.

2. Yo = ver. = 12sin35 = 6.88m/s.

3. h = 1.5 + (Vf^2 - Yo^2) / 2g,

h=1.5 + (0 - (6.88)^2) / -19.6 = 3.92m.

4. t(up) = (Vf - Yo) / g,

t(up) = (0 - 6.88) / -9.8 = 0.70s.

t(dn) = t(up) = 0.70s.

T = t(up) + t(dn) = 0.70 + 0.70 = 1.4s

= Time in flight.

d = Xo * T = 9.83 * 1.4 = 13.8m between

the girls.

- physics -
**Anonymous**, Tuesday, October 13, 2015 at 2:50pm
lll

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