Think on this. to go L meters horizontally, it has to be in the air T seconds.
in the horizontal
L=vi*cosTheta*t-1/2 g t^2 where you solve for t in terms of the other unknowns (such as vi).
now in the vertical, the final position is the same as the initial position
vf=vi*sinTheta - g t
0=vi*sinTheta-g t put in the expression you found for t above, and solve for vi.
A football is thrown upward at a(n) 37� angle
to the horizontal.
The acceleration of gravity is 9.8 m/s2 .
To throw a(n) 57.8 m pass, what must be
the initial speed of the ball?
Answer in units of m/s
1 answer