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 19 m/s at an angle 47 degrees with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

1.After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground. It is moving with a speed of 15 m/s when it reaches a maximum height of 13 m above the ground.What is the speed of the ball when it leaves Sarah's hand?

2.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

when julie throws to sarah:
Vo = 19 m/s
thetha = 47 degree

These details can be used to find the distance between them
It is nothing but range of projectile

Range = Vo^2 * sin (2*thetha) /g
= (19)^2 * sin (2*47) / 9.8
= 36.75 m

1.
When sarah throws to julie:
Let initial speed's component in vertical direction be v
Initial speed's component is horizontal direction = speed at maximum height = 15 m/s

we need to valur of v m/s

use in vertical direction:
Vf ^2= Vi ^2 + 2*a*d
Here Vf = speed in vertical direction at maximum height = 0 m/s
d = vertical distance moved by ball = 13-1.5 = 11.5 m

0 = v^2 + 2*(-9.8)*11.5
v= 15 m/s

So initial speed = sqrt (vertical componenet ^2 + horizontal compoenent ^2)
= sqrt (15^2 + 15^2)
= 21.2 m/s
Answer: 21.2 m/s

2)
time taken to get to julie = d / horizontal speed = 36.75/15 = 2.45 s
Let us calculate distance d above julie using below equation in vertical direction
d = Vi*t + 0.5*a*t^2
d = 15*2.45 + 0.5*(-9.8)*(2.45)^2
= 36.75 - 29.41
= 7.34 m
Answer: 7.34 m

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