A ball is thrown horizontal y from the roof of a building 75m tall with a speed of 4.6 m/s. What is the velocity of the ball just before it hits the ground.

Answer 1

Let’s use the downward direction as positive so that any vector that points down is no longer a negative value.

Knowns:

Vx = 4.6 m/s The x direction velocity stays constant!

Viy = 0 m/s because thrown horizontally

delta y = 75m the displacement in y-direction

a = 9.8 m/s^2 due to gravity in the y-direction

(delta y and a are downward so they are now positive)

Looking for Vfy to help us find Vf

If we are looking for something in the y direction, we can only use our y-direction variables: Viy, a, delta y. We will pick the kinematics equation that contains these 3 variables and the missing variable.

Vfy^2 = Viy^2 + 2 * a * delta y

Plug in knowns and solve for Vfy

Vfy^2 = 0^2 + 2(9.8)(75)

Vfy^2 = 1470

square root

Vfy = 38.3 m/s

Now that we have the x and y components of the velocity right before it hits the ground, we need to find the magnitude and direction of Vf. Vf is the hypotenuse of the triangle made by Vx and Vfy. We will use Pythagorean theorem to find the magnitude and inverse tangent to find the direction (the 2nd tan button on your calculator).

a^2 + b^2 = c^2

Vx^2 + Vfy^2 = Vf^2

(4.6)^2 + (38.3)^2 = Vf2

1491.16 = Vf^2

square root

Vf= 38.6 m/s Magnitude

tan^-1 (y/x) = angle

tan^-1 (38.3/4.6) = 83.2 degrees below horizontal

The velocity before it hits the ground is

Vf = 38.6 m/s at 83.2 degrees below horizontal