A baseball thrown at an angle of 55.0 degrees above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point from which it is thrown. Ignore air resistance.
Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown).
got it.... 16.0 m/s
Find the magnitude of the velocity of the baseball just before it strikes the building.
I have tried so many things it isn't even funny. HELP!
Find the direction of the velocity of the baseball just before it strikes the building.
Yeah right...like I know....
Did you set up the position and velocity vectors for the problem? That would help.
The position vector is
(x,y)=(cos(55)*v*t, (-1/2)g*t^2+sin(55)*v*t+yo)
Using the x component you see that
(1) 16m=cos(55)*v*t
For the y component we're told y_f=yo+8=(-1/2)g*t^2+sin(55)*v*t+yo
This means
(-1/2)g*t^2+sin(55)*v*t - 8 = 0
From (1) we can use 16m/cos(55) = v*t
Since you know g you can you can solve for t.
Once you know t you can determine v from (1) (Verify that I've given the correct method to do this. I see you have an answer already.)
The velocity vector for the ball is
(2) (v_x,v_y)=)=(cos(55)*v, -1g*t+sin(55)*v)
The magnitude of the velocity is the speed and it's given by
(3) Speed=sqrt((v_x)^2 + (v_y)^2)
Since you should now have the time and the inital velocity, calculate v_x and v_y and use (3) to find the speed.
Use the time you found and (2) to give the velocity vector for the ball.
If you use a diagram be sure to set up the coordinate system so you're throwing from left to right and yo is the height at the origin.
Repost if you have questions.
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