Question

A soccer ball is kicked off the top of a 12-m high building at 4 m/s. It lands 6.3 m from the base of the building 1.6 seconds later. What is the time?

Answer 1

We can use the kinematic equations to solve for the time it takes for the soccer ball to hit the ground.

Let's use the following variables:

h for the initial height of the soccer ball (12 m)

vi for the initial velocity of the soccer ball (4 m/s)

vf for the final velocity of the soccer ball (0 m/s)

d for the horizontal distance the soccer ball travels before hitting the ground (6.3 m)

t for the time it takes for the soccer ball to hit the ground (what we want to find)

Using the kinematic equation for vertical motion, we can find the time it takes for the soccer ball to reach the ground:

vf^2 = vi^2 + 2*a*h

0 = 4^2 + 2*(-9.8)*12

0 = 16 - 235.2

t_vertical = sqrt(235.2/16) = 3.05 s

So it takes 3.05 seconds for the soccer ball to hit the ground if we consider only the vertical motion.

Next, using the kinematic equation for horizontal motion, we can find the time it takes for the soccer ball to travel the horizontal distance of 6.3 m:

d = vi*t_horizontal

t_horizontal = d/vi = 6.3/4 = 1.575 s

So it takes 1.575 seconds for the soccer ball to travel the horizontal distance of 6.3 m.

Since we know the soccer ball takes the same amount of time to travel horizontally as it does to hit the ground, we can conclude that the time it takes for the soccer ball to hit the ground is 1.575 seconds.