Julie throws a ball to her friend Sarah. The ball leaves Julie\'s hand a distanc
ID: 1407685 • 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 12 m/s at an angle 47 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 8 m above the ground, and takes 1.464 s to get directly over Julie's head.
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?
Explanation / Answer
When julie throws to sarah:
time of flight = 2 vo * sin thetha / g
= 2* 12 * sin 47 / 9.8
= 1.79 s
So horizontal distance between them = vo cos thetha * time
= 12 sin 47 * 1.79
= 15.7 m
When sarah throws to Julie:
Lets horizontal componet of speed be Vx and vertical component be Vy.
time taken = 1.464 s
so, vx = horizontal distance / time
=15.7 / 1.464
= 10.7 m/s
height from where ball is thrown = 8-1.5 = 6.5 m
use:
use:
Vyf^2 = Vyi^2 - 2* g*h
0 = vy^2 - 2*9.8*6.5
vy = 11.3 m/s
speed of ball when it leaved sarah hand = sqrt (10.7^2 + 11.3^2 ) = 15.56 m/s
time taken to reach this max height be t1
Vyf = Vyi - g*t1
0 = 11.3 - 9.8*t1
t1 = 1.5 s
So it is over julie's head when it is going up.
So at t= 1.464 s, height of the ball be h1
h1 = vy*t - 0.5*g*t^2
= 11.3*1.464 - 0.5*9.8*1.464^2
= 6 m
Height above ground = 6+1.5 = 7.5 m
Very lenngth qyestion
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.