Question

A ball is thrown 190 m upward and then falls back to earth. Neglecting air resistance, how long will it be in the air?

Answer 1

38.74 s

Step-by-step explanation:

We recall the equation for the trajectory in presence of gravity (g):

y = y0 + v0 * t - 1/2 * g * t^2

When the ball is at 190 m, it's at its maximum altitude. At maximum altitude, the velocity is 0, because one instant before it's positive and one instant after it's negative.

We now calculate how much time the ball needs to get back to land, because the total time in air will be (time to reach land) * 2.

Starting from when the ball is at 190 m, the equation for the altitude is:

y = 190 m - g*t^2

We the ball reaches land, y = 0. We substitute this in the equation and we find t:

0 = 190 m - g*t^2

g*t^2 = 190 m

t^2 = 190 m / g

t =
`\sqrt{(190 m)/(9.81 m/s^2)}` = 19.37 s

Therefore, the total time in air is t * 2 = 38.74 s