Afootball quarterback is moving straight backward at a speed of 3.00 m/s when he throws a pass to a player 20.0 m straight downfield. (a) if the ball is thrown at an angle of 25.0° relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground

16.0 m/s Split the velocities into vertical and horizontal component v = sin(25)*V h = cos(25)*V Since the ball will be caught at the same height as it was released, it's upward and downward movement distances will be the same. And both will take the same amount of time. That time will be the vertical velocity divided by the acceleration due to gravity. So the total time available for the ball will be T = 2*v / 9.8 T = v/4.9 The distance the ball travels is 20.0 meters. And the time it will take is that distance divided by the velocity. So T = 20.0/h We can now set those two equations equal to each other. 20.0/h = v/4.9 Substitute the equations we have to v and h 20.0/(cos(25)*V) = sin(25)*V/4.9 And solve for V 20.0/(cos(25)*V) = sin(25)*V/4.9 98/(cos(25)*V) = sin(25)*V 98/cos(25) = sin(25)*V^2 98/(cos(25)sin(25)) = V^2 255.8598287 = V^2 15.99561905 = V Let's verify. v = sin(25)*15.99561905 = 6.76 m/s h = cos(25)*15.99561905 = 14.50 m/s T = 6.76 / 9.8 * 2 = 1.379591837 s D = 14.50 * 1.379591837 = 20.00 m So if the ball is thrown at a velocity of 16.0 m/s at an angle of 25.0° relative to the ground, it can be caught 20.0 meters away at the same height upon which it was released.