A tennis ball thrown with a velocity of 25m/s[Fwd] in the horizontal direction from the top of an 80m high building. 1. How long does it take the tennis ball to reach the ground?
2. How far is the range (how far from the base of the building will the ball land) ?
3. What is the final velocity?
Even answering one will help me out and i will give brainlest!

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Let l be the vertical distance of pendulum when it makes 25.0 degree angle to verticalcos(25.0) = l / 1.20 ml = (1.20m)(cos 25.0)so the height, h, of the pendulum above the equilibrium position is h = 1.20m - (1.20m)(cos 25.0)ep,i + ek, i = ep, f = ek, fwhen pendulum held at 25.0 degrees, initial kinetic energy is 0 jwhen pendulum passes through equilibrium point, final potential energy is 0jep, i = ek fmgh = 1/2 m v^2gh = 1/2 m v^2v = √(2gh)v = √[2 * 9.81 m/s² * ( 1.20m - (1.20m)(cos 25.0) )]v = 1.5 m/s

Answer(550) lb (1) foot each second