What must be the average speed of the person if he is to catch the ball at the bottom of the building?

Question

A ball is thrown upward from the top of a 23.1-m-tall building. The ball's initial speed is 14.9 m/s. At the same instant, a person is running on the ground at a distance of 31.3 m from the building.

Answer 1

you need to calculate how long the ball will be in the air, use the equation:

y(t)=y0+v0t-1/2gt^2 to find the height (y(t) as a function of time)

y0=initial height = 23.1m

vo=initial speed = 14.9 m/s

g=accel due to gravity = 9.8 m/s/s

and we want to find the value of t that corresponds to y(t)=0

we have:

0=23.1+14.9t-4.9t^2

this yields a quadratic equation:

4.9t^2-14.9t-23.1=0

which has the solutions -1.1 and +4.17s

so it takes 4.17s for the object to hit the ground

in order to cover 31.3 m in 4.17s, the average speed of the person will be 31.3/4.17=7.5 m/s