A ball thrown vertically up takes 4 seconds to reach its highest point. How long will it take before the ball returns from its highest point to its starting position? 4 s
4 s

40 s
40 s

1 s
1 s

10 s

4s

Explanation:

Let's analyze the vertical motion.

Let's suppose that the ball was thrown vertically with a velocity V0.

Now, the only force acting on the ball will be the gravitational force, that pulls down, then the acceleration of the ball will be equal to the gravitational acceleration:

A(t) = -g

where g = 9.8m/s^2.

For the velocity, we should integrate over time and get:

V(t) = (-9.8m/s^2)*t + v0

The maximum height will be reached when the velocity equals zero (because at this point the ball stops moving upwards).

And we know that this happens at t = 4s, then:

V(4s) = 0 = (-9.8m/s^2)*4s + v0

          v0 = 39.2m/s

Then the velocity equation is:}

V(t) = (-9.8m/s^2)*t + 39.2m/s.

Now, to get the position equation we integrate again, and get

P(t) = (1/2)*(-9.8m/s^2)*t^2 + 39.2m/s*t + p0

Where p0 is the initial position, such that:

P(0s) = p0

As we want to calculate how long will take to the ball to return to the initial point, we can assume p0 = 0m. and find for wich value of t the position equation is equal to 0m:

P(t) = 0m = (-4.9m/s^2)*t^2 + 39.2m/s*t

Let's divide by t in both sides:

         0m =  (-4.9m/s^2)*t + 39.2m/s

       (4.9m/s^2)*t = 39.2m/s

                           t =  (39.2/4.9) s = 8s

But the question is:

How long will it take before the ball returns from its highest point to its starting position?

And the ball needs 4 seconds to reach the maximum height, so the time that the ball needs to return from the max height to the initial position is:

8s - 4s = 4s.

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