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December 18, 2014

December 18, 2014

Posted by **Lauryn** on Wednesday, September 14, 2011 at 8:40pm.

a)Find the speed at which the ball clears the wall.

I calculated this correctly and got 18.13 m/s.

(b) Find the vertical distance by which the ball clears the wall.

I know that dy=vy0*t but I tried using that equation and couldn't get the answer.

I did this and it was incorrect

v0sin53 and got 18.13*sin53= 14.48

then I tried plugging that into the y, to get y=14.48sin53(2.2)-1/2(-9.8)(2.2^2)

I got 25.44-23.716=1.725, the computer assignment says that is incorrect. I don't know what I am doing wrong

(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

I can't figure out how to do the last two, can anyone please help me set up these problems so I can figure out how to solve them? Thank you.

- Physics for Engineers -
**Quidditch**, Wednesday, September 14, 2011 at 9:04pmCheck part b comments on your previous post.

- Physics for Engineers -
**Lauryn**, Wednesday, September 14, 2011 at 9:16pmthanks for your help!

- Physics for Engineers -
**Chris**, Sunday, August 31, 2014 at 9:09pmFor b.)

You got the y component correct however you multiplied it by sin53 a second time.

The correct equation should be y=18.13sin53(2.2)-4.9(4.8)

or with the y component solved for already, y=14.48(2.2)-4.9(4.8)

The ball is at height= 8.336m

It clears the wall by 8.336-6.7=1.636m

- Physics for Engineers -
**Chris**, Sunday, August 31, 2014 at 9:26pmFor c.)

Set the y component to 5.2m

5.2= 18.13sin53(t)-4.9(t)^2

solve the quadratic for t and plug into d=18.13cos53(t)

This will give you the horizontal distance until the ball lands on the roof.

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