Parachutist drops a camera

[ms. confused]

A parachutist descending at a speed of 10m/s drops a camera from an altitude of 50m. How long does it take the camera to reach the ground?

Is speed the same thing as velocity in this case? If not, what do I do? I assumed it was the same and went: Vi= 10; a=-9.8; d=50 goes into formula d=Vit + 1/2at^2. But something's wrong...

 

[quantumdude]

The camera is moving down initially, so v_i has to be negative.

 

[ms. confused]

I keep getting a negative root when I try to do the quadratic formula. WHY? Wait, I tried a different formula and now I get negative time which is impossible. Are you sure it's supposed to be -10m/s?

 

[quantumdude]

I forgot to say something: d is negative, too.

That's because d=y-y₀. If your initial height is 50 m and your final height is 0 m, you can see that d=-50 m.

 

[HallsofIvy]

First establish your coordinate system: Take the ground to be 0, positive upward. Then a= -9.8 m/s², initial velocity= -10 m/s, initial height= 50 m.
The equation for height of the camera is -4.9t²- 10t+ 50 and we want to determine when the height will be 0-
Solve -4.9t²- 10t+ 50= 0. One root will be negative- that's the time at which if you had thrown the camera up into the air it would now be at 50 m with downward velocity of -10 m/s. Obviously, you want the positive root.