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 12 m/s at an angle 49 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground. 1) What is the horizontal component of the ball's velocity right before Sarah catches it? 10 m/s s Submit You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question. 2) What is the vertical component of the ball's velocity right before Sarah catches it? s Submit You currently have O submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question 3) What is the time the ball is in the air? s Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question 4) What is the distance between the two girls? m Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question.

Explanation / Answer

1) horizontal component of velocity will not change.

vfx = v0x = 12 cos49 = 7.87 m/s

2) initial and final vertical positions are same hence vertical component will be same in magnitude and opposite in direction.

vfy = - v0y = - 12 sin49 =-9.06 m/s

3) in vertical, vfy = v0y + ay t

- 9.06 = 9.06 - 9.8 t

t = 1.85 sec

4) d = v0x t = (7.87) (1.85) = 14.5 m

5) in vertical,

0^2 - v0y^2 = 2 (-9.8) (7 - 1.5)

v0y = 10.4 m/s

in horizontal,

v0x = 14.5 / 1.454 = 9.97 m/s

speed = sqrt(v0y^2 + v0x^2) = 14.4 m/s

6) y = v0y t + ay t^2 /2

y = (10.4 x 1.454) - (9.8 x 1.454^2 / 2 )

y = 4.76 m

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