# Physics

My problem question:
"A ball is thrown up onto a roof, landing 4.0 seconds later at a height of 20.0 m above where it was thrown. The ball's velocity vector upon landing on the roof top was angled 60 degrees to the horizontal flat roof top. Find the horizontal distance that the ball traveled."

Which equation should I use to find that distance?

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1. You need more than one equation to solve this problem. From the time of flight that it takes to land on the roof, you can calculate the initial vertical component of the launch velocity, Vyo.

20.0 m = Vyo *4.0 - (g/2)*4.0^2
20 + 78.4 = 98.4 m = 4 Vyo
Vyo = 24.6 m/s
The vertical velocity component when it hits the roof is
Vy = Vyo - gt = 24.6 - 39.2 = -14.6 m/s
(It is negative because it is travelling down).

From the trajectory angle when it hits the roof,
Vy/Vx = -tan60 = -1.732
Vx = 8.43 m/s
The Vx velocity component remains constant, so it also equals Vxo at launch.

The distance traveled horizontally is Vx*4.0 s = ___

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2. 140

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