# Projectile motion of a rock

1. Sep 9, 2004

### EaGlE

A man stands on the roof of a building of height 16.3m and throws a rock with a velocity of magnitude 31.6m/s at an angle of 28.1 degrees above the horizontal. You can ignore air resistance

A.) Calculate the maximum height above the roof reached by the rock. Take free fall acceleration to be 9.80 m/s^2.

ok first i found vectors x and y at 28.1 degrees
cos(28)*31.6 = vector x
sin(28)*31.6 = vector y

x = 27.8752
y= 14.883

using this formula:
x(t) = x(0) + v(0)t + 1/2at^2
x(t) = 16.3 + (31.6m/s)t + 1/2(-9.8)t^2 <--- is that right how i set it up? is a a negative or positive number? and will there be two times? i know im doing something wrong, i just dont know what...if i solve for t now, there will be two times, which i dont know what to do with it.

B.) Calculate the magnitude of the velocity of the rock just before it strikes the ground. Take free fall acceleration to be 9.80 m/s^2

Last edited: Sep 9, 2004
2. Sep 9, 2004

### Pyrrhus

Here are some hints:

Remember,
$$V_{y} = V_{o}\sin(\theta)$$
$$V_{x} = V_{o}\cos(\theta)$$
$$V_{y} = 0$$ at max height
$$V_{x}$$ is constant (this means same value always)

9.8 must be put -9.8 because is pointing down that acceleration. And also plug in your $$V_{y}$$ value instead of just the initial $$V_{o}$$. Remember Projectile Motion it's just a mix of Free Fall Motion with Constant Velocity Motion. Read the hints!.

-Cyclovenom

Last edited: Sep 9, 2004
3. Sep 9, 2004

### EaGlE

anyway, i solved for t and got 0,6.4489s

then plugged t in the position formula agian
x(t) = x(0) + v(0)t + 1/2at^2
and got 16.30251507
but the problem only wants the part above the buliding, so i minus 16.3 from it and got .0025 is that correct? (seems incorrect)

4. Sep 10, 2004

### PureEnergy

To find the maximum height, you should only be concerned with motion in the y-direction. Physically, you want to think of how long will it take for gravity to slow down the ball until it stops. Once you find this time, you can plug it in the y analog the equation you have in part a to find the maximum height.

In part b, you want to first calculate the y component of velocity as it hits the ground. When the rock is thrown upward and then comes back down to the initial height (16.3m), you should recognize the velocity in the y-direction (vy) at that point has the same numerical value as the initital vy but is negative since it's heading downward. You can use the following equation to find the final vy:

$$v_{y}^2 = v_{yo}^2 + 2.0 a_{y}(y-yo)$$

Now, using vx which doesn't change throughout the motion, you can now solve for the velocity magnitude

5. Sep 10, 2004

### Pyrrhus

Ok let me give you a little help.

$$Y_{max} :$$
$$V_{y} = 0$$
So using this equation:
$$V = V_{o} + at$$
$$0 = V_{y} - gt_{max}$$
$$t_{max} = \frac{V_{y}}{g}$$

Now using this equation:
$$Y = Y_{o} + V_{o}t + \frac{1}{2}at^2$$
$$Y_{max} = Y_{0} + V_{y}t_{max} -\frac{1}{2}gt_{max}^2$$

Combining Equations:
$$Y_{max} = Y_{0} + V_{y}\frac{V_{y}}{g} - \frac{1}{2}g(\frac{V_{y}}{g})^2$$

$$Y_{max} = Y_{0} + \frac{V_{y}^2}{g} -\frac{1}{2}\frac{V_{y}^2}{g}$$

-Cyclovenom

Last edited: Sep 10, 2004