Posted by **leonard** on Friday, October 4, 2013 at 10:27am.

A person is riding on a flatcar traveling at a constant speed v1= 15 m/s with respect to the ground. He wishes to throw a ball through a stationary hoop in such a manner that the ball will move horizontally as it passes through the hoop. The hoop is at a height h=4 m above his hand. He throws the ball with a speed v2= 18 m/s with respect to the flatcar. Let g=10 m/s2 and neglect air drag completely

When the ball leaves his hand, what is the direction of the velocity vector of the ball as seen from the ground? (angle αground with respect to the horizontal in degrees

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**Elena**, Friday, October 4, 2013 at 3:07pm
The ball moves in projectile motion. When it is moving horizontally, v(y) =0

+x is directed to the right, +y is directed upward, a(x)= 0, a(y) =-g

v₀(y)=sqrt(2gh) = sqrt(2•9.8•4) = 8.85 m/s

v= v₀(y)-gt

At the top point v=0 =>

t= v₀(y)/g =8.85/9.8 =0.9 s.

The horizontal component of the ball’s velocity relative to the man is

sqrt{v₂²-v₀(y) ²} = sqrt{18²-8.85² } =15.67 m/s:

the horizontal component of the ball’s velocity relative to the hoop is

15.67 +15 =30.67 m/s,

and the man must be 30.67•0.9 = 27.6 m in front of the hoop at release.

Relative to the flat car, the ball is projected at an angle

tanα = v₀(y)/15.67 =8.85/15.67=0.565

α=29.46⁰

Relative to the ground the angle is

tanβ =8.85/(15.67 +15) =8.85/30.67= 0.289

β=16.1⁰

In both frames of reference the ball moves in a parabolic path.

The only difference between the description of the motion in the two frames is the horizontal component of the ball’s velocity.

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**Anonymous**, Sunday, October 6, 2013 at 7:34am
Thanx a lot

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**pamporis**, Monday, October 7, 2013 at 7:05am
how did you find distance x?

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**xyz**, Monday, October 7, 2013 at 10:50am
thank u very much elena ..that was a great help..! :)

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