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Projectile Motion-Shooting a Ball

by atbruick
Tags: ball, motionshooting, projectile
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atbruick
#1
Sep20-10, 03:46 PM
P: 20
1. The problem statement, all variables and given/known data
A ball is shot from the top of a building with an initial velocity of 15 at an angle = 45 above the horizontal. What are the x and y components of the initial velocity? If a nearby building is the same height and 59 m away, how far below the top of the building will the ball strike the nearby building?




2. Relevant equations
Trig equations of sin and cos
sinTheta=opp/hyp cosTheta=adj/hyp


3. The attempt at a solution
I thought the x component at initial velocity would be zero, and the y component was sin45=y/15, which I got to be 11; Either one of these or both are wrong, my homework doesnt tell me if one is right or not. Not sure how to approach this because I thought the sin for y was correct.
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paco_uk
#2
Sep20-10, 05:25 PM
P: 22
If the x component of the initial velocity was zero then the ball would be travelling straight upwards. How have you calculated this?
atbruick
#3
Sep20-10, 08:08 PM
P: 20
oh opps so just like you calculated y by using sine use cosine to get x. Thank you! I also just tried to find the second part, the distance it hits below the roof; I needed to find t so I used X=Xo+Vxot, 59=0+11t, and got t to be 5.36. Then plugged that into Y=Yo+Vyot-.5at^2, and got Y=0+11(5.36)-.5(-9.80)(5.36^2) and got 200 which was wrong. What am I plugging in wrong in the equation?

paco_uk
#4
Sep21-10, 04:45 AM
P: 22
Projectile Motion-Shooting a Ball

I think you are using all the right equations but you need to be careful with minus signs.

If you are regarding upwards as positive (a good idea) then your expression for y should read:

[tex]

Y=Y_0+V_y(0)+\frac{1}{2} a t^2

[/tex]

Then you can consider in which direction acceleration due to gravity points.

Also watch out for rounding errors. You might find that you need more decimal places in your value for Vx and Vy.


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