# Physics

posted by Kyle

A man stands on the roof of a 15m tall building and throws a rock with a velocity of magnitude 30m/s at an angle of 33deg above the horizontal. Ignore air resistance. Calculate:

a) the maximum heigh above the roof reached by the rock
b) the magnitude of the velocity of the rock just before it strikes the ground
c) the horizontal range from the base of the building to the point where the rock strikes the ground

My work so far:

a) I found the initial x and y components to be Vx = 25.16 and Vy = 16.33

Then I found the time it took for the rock to reach its heightest point:
Vf = Vi + a*t
0 = (16.33) + (-9.8)(t)
t = 1.6s

then I plugged that in to find the height

H_max = [(16.33) + 0]/2 * (1.6) = 13.55 m

b) here is where I'm a little lost

I set up this equation to find the time it takes for the rock to go through its entire fall from the building's top to the ground:

y = vy*t + 1/2*g*t
-15 = (16.33)*t - 4.9*t^2

Then I used the quadratic equation to find the time is t = 3.98s

I plugged that back into the Vf = Vi + a*t equation to find the final velocity

Vf = Vi + a*t
Vf = (16.33) + (-9.8)(3.98)
Vf = -22.67

magnitude is the absolute value of this = 22.67 m/s

But my solution book says that the answer is 34.6 m/s so what part did I do wrong?

c) I used 3.98s to plug into
x = (25.16)*(3.98)
x = 100.64 m

Am I doing the problem correctly? I'm fairly sure that I am for the most part.

1. tchrwill

a) h = V^2(sin^2(33))/2(9.8)

b) V = 33m/s (no atmosphere)

c)Determine time "t" to fall from building roof height to ground from 15 = 16.33t + 9.8t^2/2
Horizontal distance traveled from building base to impact on ground is
d = V^2(sin(2µ))/g + 25.16t

2. jesus

you need to make your g positive and keep the initial position negative to get the correct t (.704s)

then that gives you your Voy.(21.9)
now the speed = sqt (25.16^2+21.9^2)=33.97
i guess the difference between the book answer and this one is sig figs

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