You throw a ball at 37° above the horizontal and you want to hit a target on the ground, 20 ft from where you are standing. how fast should you throw the ball? what will be its final velocity?

Let the launch velocity be V ft/s.

The horizontal component of velocity is
Vx = Vcos(37) = 0.7986V ft/s

The vertical component of velocity is
Vy = Vsin(37) = 0.6018V ft/s

The time, t, to reach maximum height is determined by
0 = Vy - gt0 = (0.6018V ft/s) - (32.174 ft/s^2)(t s)
0 = 0.6018V - 32.174t
t = (0.6018V)/32.174 = 0.0187V s

The time to travel 20 ft horizontally takes a time of 2t. Therefore
Vx*(2t) = 20
(0.7986V)*(2*0.0187V) = 20
0.0299V^2 = 20
V^2 = 668.8963
V = 25.86 ft/s

The final velocity will have components (Vx,  Vy). Its magnitude will be V or 25.86 ft/s.

V = 25.9 ft/s (nearest tenth)

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