How fast should the body be thrown down from a height of 200 m so that it falls to the ground in 5 s?

Let’s point the y-axis down. The origin is at the point of throwing the body.

The change in the y coordinate with time t will look like this:

y = v0t + gt ^ 2/2, where v0 is the initial velocity, g is the acceleration of gravity.

At the end of the fall, the value y = 200 m, t = 5 s.

200 m = v0 * 5 s + (10 m / s ^ 2 * 25 s ^ 2) / 2;

200 m = v0 * 5 s + 125 m;

75 m = 5 s * v0.

Answer: v0 = 15 m / s.




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