and 1.91 seconds later the ball falls into the net at an angle of 60.0 degrees f

Question

and 1.91 seconds later the ball falls into the net at an angle of 60.0 degrees from the horizontal. The ball enters without hitting the blackboard. The ball was releases a horizontal distance of 10.0m from the basket. In this problem, treat up as positive and down as negative. Assume air resistance is negligible.

1) What is the x-component of the balls velocity at the moment it enters the net?

2) What is the y component of the ball's velocity at the moment it enters the net? Indicate the direction with a plus or minus sign.

3)What is the y component of the ball's intial velocity. at t=0

4)What are the magnitude and angle with respect to the horizontal of the the ball's inital velocity?

5) If the height of the basket is 3.05 meters, what is the inital height of the ball?

I think i have 1 being 2.615 but using sin to find 2 I get -14.2 which is cool but then for 3 I get the same answers. As for the rest I have no clue.

Explanation / Answer

1) vx = 5.2356 m/s.   v = d/t = (10 m)/(1.91 s) = 5.2356 m/s. This is because the x-component of velocity is constant, gravity does not affect it.

2) Since y/x = tan , vy = vx tan = (5.2356 m/s)(tan -60) = -9.0683 m/s

3) You can treat the y-component independently from the x-component, so this just becomes an up-and-down problem. Since vf = vi + gt, solving for vi:

vi = vf - gt = -9.0683 m/s - (-9.81 m/s2)(1.91 s) = 9.6688 m/s

4) magnitude is sqrt(vx2 + vy2) = sqrt(5.2356 m/s)2+(9.6688 m/s)2) = 10.9953 m/s

angle is arctan(vy/vx) = 61.55 degrees

5) Easiest to treat as energy problem and just look at vertical component:

gh = vf2 - vi2

h = (vf2 - vi2)/g = ((9.0683 m/s)2 - (9.6688 m/s)2)/(9.81 m/s2) = -1.1468 m

so the initial height is 3.05 - 1.1468 = 1.902 m

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