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 11 m/s at an angle 51 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, and is moving with a speed of 9 m/s when it reaches a maximum height of 7 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

I get 15.321 m for the distance. Close enough. One of us is more precise than the other but the approach is most likely the same. At the maximum height, the y component of velocity is zero, so the only component left is the x component, which is constant though the duration of the flight: If Vo is the initial velocity that you're looking for, and a is the angle with the ground 10 = Vx = Vo * cos a If the time for the second flight is t, The total x distance is Vo * (cos a) * t = 15.32 (from the first part of the question) Substitute Vo*cos a = 10 and solve for t: t = 15.32/10 = 1.532 The total y distance for the second flight is Vo * (sin a) * t - (1/2)*(9.8) * t^2 = 0 divide both sides by t: Vo * sin a - 4.9t = 0 add 4.9t to both sides: Vo * sin a = 4.9t plug in t = 1.532 Vo * sin a = (4.9)*(1.532) = 7.5068 I want to do something with the trig identity (sin a)^2 + (cos a)^2 = 1 (Vo*cos a)^2 + (Vo*sin a)^2 = 10^2 + 7.5068^2 = 156.35 = (Vo)^2*[ (cos a)^2 + (sin a)^2 ] = Vo^2 Vo = sqrt(156.35) = 12.504

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