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Home Q & A Board Science-Related Homework Help Physics
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rjframe rjframe
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11 years ago
Here is the problem: A ball is launched from the ground and lands 40.0m away from the launch point, 2.44 sec. later.  Find the magnitude of the initial velocity vector and the angle it is above the horizontal.  

I have the answers [20.3 m/s and 36.1 degrees (respectively)] but I'm totally lost on how to go about solving it!! Any help would be greatly appreciated. Thanks!
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#1
11 years ago
Suppose the velocity is u and it is launched at an angle theta.
So u*cos(theta)*2.44s=40m and
0=(u*sin(theta)*2.44)-(0.5*g*2.44^2) . (here the formula being s=ut+1/2 at^2)
you get two equations and two unknowns . solve them
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