After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah\'s

Question

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 14.0 m/s when it reaches a maximum height of 7.0 m above the ground.

What is the speed of the ball when it leaves Sarah's hand? Vy = 17.43301466 m/s

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

Displacement = 18.07364114m

I used t = displacement/Vx, then used y= yinitial + Vyt - (g/2)t^2

I got 15.83082939 m, but it's wrong.

Explanation / Answer

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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