# Finding Projectiles' Flight Times...

1. Feb 20, 2017

### EthanVandals

1. The problem statement, all variables and given/known data
Two projectiles are launched from the ground with initial speeds of 100 meters per second. One is launched straight up, the other at a 30-degree angle. How much time elapses between each's return to ground level?

Theta = 30
g = 10 m/s
Vinitial = 100

2. Relevant equations
y = vt - 1/2gt^2

3. The attempt at a solution
As you can see in the image, I attempted to use the equations that I THINK were correct for this problem. My professor just threw a bunch of equations up on the board, some of which he said we don't need for now and they're just there to be there...so, hopefully I did these right.

2. Feb 20, 2017

### PeroK

The attachment doesn't make your approach clear. How did you go about solving this?

And, what answer did you get?

3. Feb 20, 2017

### EthanVandals

I wrote the formula that I used on the paper...it's right underneath the given variables. My answers were that the object thrown straight up in the air took 20 seconds to fall, and the object that was thrown at a 30 degree angle took 5 seconds to fall.

4. Feb 20, 2017

### PeroK

Why 5 seconds?

5. Feb 20, 2017

### EthanVandals

Because, after plugging in 0 for y (when the object hits the ground), I got the equation 0 = 50t-10t^2, and after solving for t, I got 5.

6. Feb 20, 2017

### PeroK

So, why 20 seconds in the first case?

7. Feb 21, 2017

### PeroK

Don't you think the difference between 20 and 5 is too large?

8. Feb 21, 2017

### Arif Setiawan

You can view from vertical perspective.. from ground to max for each projectile.

First, vy=100.. using vt=v-gt
t=v/g=100/10=10s

Second, vy=100.sin30=50 m/s
t=5seconds

Total time for projectile 1 is 20s
Time for projectile 2 is 10s
So difference/time elapse is 20-10=10s