Question

A ball is thrown with a horizontal velocity of 80 m/s from the top of a 250 m high building. How long will it be in the air?

Answer 1

Answer:

19.0 s

Explanation:

Since the motion of the ball is a free-fall motion, we can solve the problem by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where, chosing upward as positive  direction, we have:

s = -250 m is the displacement of the ball to reach the ground

u = +80 m/s is the initial velocity of the ball

t is the time

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Re-writing the formula we have:

$-250 = 80t -\frac{1}{2}(9.8)t^2\\4.9t^2-80t-250=0$

This is a second-order equation, so we solve it by applying the usual formula to find t:

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-80)\pm \sqrt{(-80)^2-4(4.9)(-250)}}{2(4.9)}$

And considering only the positive solution, from the calculation we get

t  = 19.0 s

So, the ball will be in air for 19.0 s.