Standing on top of a building, Bob tosses a baseball straight up with an initial speed of 15 m/s. It takes 4.0 s for the ball to hit the ground. What is the vertical distance of the building? (g = -9.8 m/s²)

48.26 m

Explanation:

time to goes up (till stop for a while in the air - maximum height)

vt = vo + a t

0 = 15 + g . t

0 = 15 + (-9.8) . t

9.8t = 15

t = 1.531 s

so the time left to goes down is

4.0 - 1.531 = 2.469 s

height from the top of building can find it by using

vo =√(2gh)

15 = √(2)(9.8).h

15² = 19.6h

h = 225/19.6 = 11.48 m

so the distance of maximum height to the ground is

t = √(2H/g)

2.469 = √(2H/9.8)

2.469² = 2H/9.8

6.096 = 2H/9.8

2H = 6.096 x 9.8 = 59.74 m

so the vertical distance of the building (or the building height's is)

H - h = 59.74 - 11.48 = 48.26 m

the average speed is 24.57 km/hr