# Kinematics motion, projectile. This was an mcat problem. Thanks

1. Jun 29, 2013

### HumorMe81

A firecracker with mass 100g is propelled vertically with a launch force of 1.2N applied over 5s, after which it explodes into two 50g fragments. One fragment goes horizontally at 15m/s and the other goes to the right at angle 53 degrees with the horizontal at speed 25 m/s. find the distance between them once they hit the ground.

This is what I did:

Force up - Fg = ma
1.2N - 1N = .1kg a
a = 2m/s^2 upward

I used this info to calculate the speed right before the explosion:
V = Vi + at
V = 0 + 2m/s^2 * 5s= 10m/s
The distance elapsed in this time is:
Y = Vi*t + 1/2gt^2
Y = 1/2 (2m/s^2)*(5s)^2 = 25m

Since the problem says after the 5s the cracker explodes, I thought the fragment that goes left takes 5s to come back down and the distance it would travel is 75m. But I'm wrong. It travels 33m.
X = Vi*t + 0

The fragment moving to the right is at an angle so it has a horizontal and vertical components. Vx is 15m/s and Vy is 20m/s.

Can someone guide me please? Thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 29, 2013

### Staff: Mentor

You need to determine the time it takes to hit the ground. What's the acceleration of any falling body?

Same thing for this one. How long does it take before it hits the ground.

Hint: The time it takes to hit the ground depends only on the vertical motion.

3. Jun 29, 2013

### rude man

Your a is off by 10%.
Why compute this?
EDIT:
Correct this with the right a. The acceleration is a, not g as your formula states, but I see you did use a instead of g after all. So you're not off by much for Y.

I see doc al has put in so I'll leave it to him for the rest.

Last edited: Jun 29, 2013
4. Jun 29, 2013

### Staff: Mentor

The MCAT may well use g = 10 m/s^2 for ease of calculation.

5. Jun 29, 2013

### HumorMe81

Ok, so i figured out what I was doing wrong. The fragment moving left travels a vertical distance of 25m. Right before it starts coming back down, its velocity at the highest point would be 0. I used this info to calculate time on the downward flight:

Y = Vi*t + 1/2gt^2
25 = 0 + 1/2(10m/s^2)*t^2
t = 2.2s

Now i can find the displacement of the fragment moving left:

X = Vi*t = 15m/s * 2.2s = 33m

For the fragment moving to the right, here's what I attempted:
After the explosion, vertical velocity is 20m/s and at the highest vertical distance it would be 0. The only acceleration it has during this segment is due to gravity downwards. I used this info to calculate the vertical height of the segment and time:

Vf = Vi + at
0 = 20m/s + (-10m/s^2)*t and t= 2s

Y= vi*t +1/2gt^2
Y = 20m/s *2s + 1/2 (-10m/s^2)(2s)^2
Y = 20m

The total vertical height is then 45m. Assuming velocity at the highest peak when fragment starts falling down to be 0m/s, i attempted to calculate the time of flight down:

Y = Vi*t + 1/2gt^2
-45m = 0 + 1/2(-10m/s^2)(t)^2
t turns out to be 3s and i used it to calculate horizontal displacement:

X = Vi*t = 15m/s * 3s = 45m
So the horizontal distance between the two fragments once they hit ground is 33m + 45m = 78m
However, that's not the right answer :(. What am I doing wrong?

6. Jun 29, 2013

### Staff: Mentor

For the particle going right you need the total time after the explosion, not just the time from the highest point.

7. Jun 29, 2013

### HumorMe81

Oh, so the time is 5s then. 2s after the explosion and 3s to come back down? If the total time is 5s then horizontal distance is 15m/s*5s = 75m and the total distance between the two fragments after they hit the ground is 33m+75m = 108m. That's the answer listed in the book.