Calculating height (projectile motion)

[roam]

Homework Statement

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?

The correct answer is 23.5m.

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?

 

[rl.bhat]

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