Julie throws a ball to her friend Sarah. The ball leaves Julie\'s hand a distanc
ID: 1471139 • Letter: J
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 10 m/s at an angle 36 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
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 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?
6)
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
(3) The maximum Height of the projectile
H = (VSin(theta))2/2g = (10Sin36)2/(2*9.81) = 1.76 m
Height above the ground = 1.76+1.5 = 3.26 m
(4) The distance between the two girls
R = V2Sin2(theta)/g = 102Sin(2*36) / 9.81 = 9.694 m
(5) Maximum height above the gound = 5 m
therefore the maximum height of the projectile (H)= 5-1.5 = 3.5 m
Let say the initial velocity is V therefore
horizontal velocity is VCos(theta) = 11 m/s - -------------(1)
using H = (VSin(theta))2/2g
3.5 = (VSin(theta))2/2g
VSIn(theta) = 8.286 m/s --------------(2)
Using 1 and 2
we get
V = 13.77 m/s
(6)
To cover the distance between two = 9.694 m
time requires is = 9.694/11 = 0.88 s
Height in this time
H = VSin(theta)*t - 0.5gt2
H = 8.286*0.88 - 0.5*9.81*0.882
H = 3.49 m
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