1) What is the horizontal component of the ball’s velocity right before Sarah ca

Question

1) What is the horizontal component of the ball’s velocity right before Sarah catches it? (13.78)

2)What is the vertical component of the ball’s velocity right before Sarah catches it? (11.5)

3) What is the maximum height the ball goes above the ground? (8.32)

4)What is the distance between the two girls? 32.3

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 15 m/s when it reaches a maximum height of 9 m above the ground.

What is the speed of the ball when it leaves Sarah's hand? 19.28

i just need this

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

Explanation / Answer

a) There is not force in the X direction

so Vx = 18* cos40degree = 13.78 m/sec

2) The vertical component is Vy = 18 * sin40degree = 11.57 m/sec

3) v = Vy + at

0 = 11.5- 9.81 t

  11.5 /9.8 = t

t = 1.179

x = x0+v0t+1/2 at^2

= 1.5 +11.57* 1.179 +0.5*9.81* 1.179*1.179 = 8.32

after solving 4) we find out Range and t ?

so range = sin80 * V^2 / g = sin80 * (18)^2 / 9.8 = 32.52m

and t = range/ Vx = 32.52/13.78 = 2.35 sec

4) with an uniform horizontal velocity . this is the distance the ball travels in 2.35 sec

. so distance x = Vx * t = 13.78 *2.35 = 3.23m

5) It reahces a height of ( 9-1.5) = 7.5 m above release height.

so vertical component of its velocity = sqrt (2gh) = 12.12 m/sec

The ball's horizontal component of velocity of 15m/sec so vertical velocity component at max height.

Velocity leaving sarah's hand = sqrt ( (12.12)2 + (15)2 ) = 19.28 m/sec

6) y(t) = H +Vyt -1/2 g t^2

= 1.5 + 11.57*2.35 -0.5 *9.8 * 2.35*2.35 =1.65 m

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