# Maximum height of a projectile thrown from a rooftop

1. Jan 28, 2014

### s.dyseman

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

A man stands on the roof of a building of height 14.6m and throws a rock with a velocity of magnitude 30.8m/s at an angle of 33.2∘ above the horizontal. You can ignore air resistance.

Calculate the maximum height above the roof reached by the rock.

2. Relevant equations

Velocity and position equations

Basic trigonometry

3. The attempt at a solution

Initially, I solved for the y-component of the velocity vector given: V=30.8*Sin(33.2)=16.86m/s

Then, I solved for the amount of time it would take for the rock to reach maximum height, where the velocity of the y-component vector is equal to 0: Vy=Voy+g*t=16.86-9.8t=1.72s

I plug this time into the position equation of Y=Yo+Voy*t+g*t^2=14.6+16.86(1.72)-4.9(1.72)^2=29.1m

So, the maximum height should be equal to 29.1m. Not sure why this is incorrect... Perhaps I calculated the vector incorrectly?

2. Jan 28, 2014

### Staff: Mentor

You calculated the maximum height above the ground (and I can confirm this value).

3. Jan 28, 2014

### s.dyseman

Ah, so I just needed to subtract the height of the roof... Simple detail I missed... Thank you!