# Projectile Motion Analyzed

1. Feb 10, 2013

### Blox_Nitrates

1. The problem statement, all variables and given/known data
A projectile is fired horizontally from a gun that is 45.0m above flat ground, emerging from the gun with a speed of 250m/s. How long does the projectile remain in the air?

2. Relevant equations
unsure of what equation to use.

3. The attempt at a solution
I tried to use t = Δx/Vox which didn't work because it gave me 45/250=.18

*The answer is 3.03 seconds in the back of the book.

Last edited: Feb 10, 2013
2. Feb 10, 2013

### CAF123

If you want to use that eqn to find t then you will need the horizontal distance that the body travels. What other kinematic equations do you know?

3. Feb 10, 2013

### Blox_Nitrates

I have these
v=vo+at
x-xo=vot+(1/2)at^2
v^2=Vo^2+2a(x-xo)
x-xo=(1/2)(vo+v)t
x-xo=vt-(1/2)at^2

Sorry about asking something probably simple as to I'm new to Calculus based Phyics and have never seen physics before.

4. Feb 10, 2013

### LawrenceC

Question: If you dropped a bomb from a stationary hot air balloon while simultaneously dropping another bomb from a horizontally flying aircraft at the same altitude, would there be any difference in the amount of time it takes for the bombs to hit the ground?

5. Feb 10, 2013

### Blox_Nitrates

No, there wouldn't be.

6. Feb 10, 2013

### LawrenceC

That is correct. Therefore you have a free fall problem. The muzzle velocity is analogous to the moving airplane. It is extraneous data that is supplied to confuse the student.

7. Feb 10, 2013

### Blox_Nitrates

So in this case t = (V-Vo)/a = ((250m/s)/-9.8m/s^2)?

but a = 0 due to no acceleration because of horizontal motion.

Last edited: Feb 10, 2013
8. Feb 10, 2013

### LawrenceC

No, the 250 m/s is the horizontal velocity and is extraneous data. The problem is the same as if you dropped a brick off a 45 m high building. How long would it take to reach the ground?

9. Feb 10, 2013

### Blox_Nitrates

t = (0-45m/s)/-9.8m/s^2 = 4.59s?

Is that correct?

10. Feb 10, 2013

### LawrenceC

You are attempting to subtract meters/second from meters. You cannot do that.

What about the formula

y = Vo*t + .5*a*t^2

where Vo is initial VERTICAL velocity
a is acceleration of gravity
t is time
y is distance of free fall

11. Feb 10, 2013

### Blox_Nitrates

45=0*t+.5(9.8)t^2

(45/4.9)^(1/2) = 3.03s

Thank you so much for your help LawrenceC!

So if I was asked At what horizontal distance from the firing point does it strike the ground? I would use a similar eqn to solve it while plugging in t correct?

Last edited: Feb 10, 2013
12. Feb 10, 2013

### Staff: Mentor

The horizontal velocity does not change (neglecting air drag), so you can use your initial formula (in post 1) to calculate that distance.

13. Feb 10, 2013

### Blox_Nitrates

I'm not sure if this is correct but I used Savg=TotalDistance/Δt
Total Distance = 250m/s(3.03s) = 758m rounded up.
The answer does match the one in the book.

14. Feb 10, 2013

### Staff: Mentor

That is correct.

15. Feb 10, 2013

### Blox_Nitrates

This question throws me off.
A projectile's launch speed is five times its speed at maximum height. Find launch angle θo.

16. Feb 10, 2013

### Staff: Mentor

That is a nice question.

At maximum height, can you determine the direction of motion? Can you say something about the launch velocity components, based on that?

17. Feb 10, 2013

### Blox_Nitrates

I can say that the max height will be vertical and the direction of motion is upwards. Wouldn't that be a 90 degree angle of (pi/2)?

Last edited: Feb 10, 2013
18. Feb 10, 2013

### Staff: Mentor

If it would go upwards there, it would not have reached its maximal height.

It is pi, pie is something to eat. And it depends on the definition of the angle.

19. Feb 10, 2013

### Blox_Nitrates

Sorry. So would the maximal height be somwhere between 45-90 degrees?

20. Feb 10, 2013

### haruspex

It doesn't mean that the angle is chosen to achieve maximum height. The angle is whatever it is. mfb is asking what the direction of motion of the projectile will be when the projectile is at its maximum height on the trajectory. Sketching it might help.