# Basic projectile question help?

1. Mar 5, 2014

### rexorsist

I know how to do most projectile questions, but if just can't seem to get the correct answer for this:

6) A helicopter is ascending vertically with a velocity of 8.0 m/s at a height of 120 m when a package is dropped out of the door. How much time passes before the package hits the ground?

I know that:

Displacement: 120m
Gravity: 9.8 m/s^2

I don't see how the velocity of the helicopter is relevant to the time that it takes the package to hit the ground.

Can someone point me to the right direction? I have a huge unit test tomorrow and need help.

2. Mar 5, 2014

### jackarms

Well, your setup seems to imply that the package starts at rest when it is dropped, but are you sure about this? Before the package is dropped, what is it's velocity inside the helicopter?

3. Mar 5, 2014

### rexorsist

Oh, of course! The velocity of the helicopter would be the initial velocity for the package (but negative). This means I'd first have to find the final velocity (using final velocity^2-initial velocity^2=2xgravityxdisplacement) and then use that to find time (with V2=V1+gravityxtime).

I got the correct answer, but apparently my teacher says there is a short cut and that this question can be done in one step rather than two. Is there any way to combine the motion equations to find the time in one step (without having to find final velocity first?).

4. Mar 5, 2014

### rexorsist

Oh, of course! The velocity of the helicopter would be the initial velocity for the package (but negative). This means I'd first have to find the final velocity (using final velocity^2-initial velocity^2=2xgravityxdisplacement) and then use that to find time (with V2=V1+gravityxtime).

I got the correct answer, but apparently my teacher says there is a short cut and that this question can be done in one step rather than two. Is there any way to combine the motion equations to find the time in one step (without having to find final velocity first?).

5. Mar 5, 2014

### jackarms

You could use this equation:
$$\Delta y = v_{0}t - \frac{1}{2}gt^{2}$$
Although it would involve solving a quadratic, which would be slightly more involved.