# Calculating height (projectile motion)

1. Mar 31, 2010

### roam

1. The problem statement, all variables and given/known data

A rock is thrown from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53 degrees above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. How tall is the building?

3. The attempt at a solution

$$v_{ix}=12.2 cos 53 =7.34$$
$$v_{iy}=12.2 sin 53 = 9.74$$

$$t=\frac{v}{x}=\frac{7.34}{25}=0.29$$

$$y=v_{iy}+1/2gt^2=0.29 \times 9.74 + \frac{1}{2}(9.81)(0.29)^2 = 3.23$$

So why is my answer wrong?

2. Mar 31, 2010

### rl.bhat

Your formula for y is wrong. It should be
y = viy*t - 1/2*g*t^2.