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:

Question

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

Answer 1

Magnitude is not the absolute value of the velocity. Magnitude is the combination of velocity in the x and y directions. You have to use Pythagorean Theron to solve for magnitude.

For this problem it would be...
Vf^2=22.6^2 + 30^2

Answer 2

Kyle,
Thanks for solving thw first part
V=√vx^2+vy^2

Answer 3

a. 13.6m
b. 34.6m/s
c. 103m

Answer 4

b) solution is like this.
Y=vyt-1/2gt^2 --------------(i)
15=16.33t-5t^2 (g=10m/s)
Use quadratic method.find t
The value of t comes 4.01s...

Again use euqation (i)
To find vy ,where t=4.01s.
Then vy comes 23.79m/s.

Now,
for magnitude.
V^2=(25.16)^2+(23.79)^2
: .V=34.6m/s

Answer 5

At first to find time use (-15).
Then to find vy use just (+15).