Magnitude of the Velocity- Projectile Problem

[haengbon]

Homework Statement

A man stands on the roof of a building that is 30.0 m tall and throws a rock with a velocity of magnitude 60.0 m/s at an angle 33 degrees from the horizontal. calculate (a) the maximum height above the roof reached by the rock. (b) the magnitude of the velocity of the rock just before it strikes the ground.

here's the figure :D

http://img245.imageshack.us/img245/7768/doc1.png

Homework Equations

V^2 = Vo^2 + 2gd
V= Vo + gt
Tan (teta) = Vx/Vy
We use 10 for gravity, not 9.8.

The Attempt at a Solution

Vox = 60cos33 = 50.32 m/s
Voy = 60sin33 = 32.67 m/s

(a) V^2 = Vo^2 + 2gd
0 = (32.67)^2 + 2(-10) d
hmax = 53.39 m

I got maximum height (hmax) correct :D but I'm having a hard time with magnitude of the velocity T T

total time of travel to hmax

V=Voy + gt
tmax= 3.268 s

total time of travel

tmax x 2
t = 6.54 s

since: tan (teta) = Vy/Vx

Vx = Vox = 50.32 m/s
Vy = ?

Vy = Voy - gt
= 32.67 - 10 (6.54)
Vy = -32.73 m/s

tan (teta) = Vy/Vx

teta = tan^-1 | Vy |/| Vx |

= tan^-1 |-32.73 m/s| / | 50.32 m/s |


teta= 33.04 degrees

V = SQUARE OF (Vx^2 + Vy^2)
= SQUARE OF ( 50.32^2 + -32.73^2)
= SQUARE OF ( 3,603.3553
V = 60.027 m/s

therefore :

60.027 m/s, 33.04 degrees, 2nd quadrant.

is this the correct answer? :D this is what I solved but I'm not confident if it's really correct :) please can someone help ?

 

[rl.bhat]

Your hmax is from the top of the building. To this add height of the building. At that point initial velocity is zero. Find the time taken by the rock to reach the ground. And the velocity vy with which it reaches the ground.

 

[kerimek]

Yes, your approach and calculations are correct. The magnitude of the velocity just before it strikes the ground is 60.027 m/s. Good job!