A ball thrown horizontally at 12.6 m/s from the roof of a building lands 20.0 m from the base of the building How long does it take for the ball to reach the ground and How tall is the building?

1.59 seconds

12.3 meters

but if you are wise you will read the entire answer.

Explanation:

This is a good question -- if not a bit unusual. You should try and understand the details. It will come in handy.

Time

Given

a = 0 This is the critical point. There is no horizontal acceleration.

d = 20 m

v = 12.6 m/s

Formula

d = vi * t + 1/2at^2

Solution

Since the acceleration is 0, the formula reduces to

d = vi * t

20 = 12.6 * t

t = 20 / 12.6

t = 1.59 seconds.

It takes 1.59 seconds to hit the ground

Height of the building

Givens

t = 1.59 sec

vi = 0     Another critical point. The beginning speed vertically is 0

a = 9.8 m/s^2   The acceleration is vertical.

Formula

d = vi*t + 1/2 a t^2

Solution

d = 1/2 a*t^2

d = 1/2 * 9.8 * 1.59^2

d = 12.3 meters.

The two vi's are not to be confused. The horizontal vi is a number other other 0 (in this case 12.6 m/s horizontally)

The other vi is a vertical speed. It is 0.

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14