A photographer in a helicopter ascending vertically at a constant rate of 1.3 m/s accidentally drops a camera out the window when the helicopter is 40.0 m above the ground. How long will it take the camera to reach the ground?
Which equation should I use?
a) Y=Yo + Vo(t) + (1/2)(-g)(t^2)
b) V=Vo + (-g)(t)
c) (V^2)=(Vo^2) + 2(-g)(Y-Yo)
d) Y-Yo=(1/2)(V-Vo)(t)
To solve this problem, you can use the equation for vertical motion under gravity, which is a) Y=Yo + Vo(t) + (1/2)(-g)(t^2).
Let's break down the equation:
- Y represents the final position of the camera (height above the ground).
- Yo represents the initial position of the camera (height when dropped, which is 40.0 m above the ground).
- Vo represents the initial vertical velocity of the camera (velocity at the moment it was dropped, which is 0 m/s because the camera was not moving vertically).
- t represents the time it takes for the camera to reach the ground.
- g represents the acceleration due to gravity (approximately 9.8 m/s^2).
Since the camera was dropped from rest, Vo = 0, and the equation simplifies to Y = Yo + (1/2)(-g)(t^2).
Now plug in the known values:
Y = 0 (since the camera reaches the ground, Y = 0).
Yo = 40.0 m.
g = 9.8 m/s^2.
The equation becomes:
0 = 40.0 + (1/2)(-9.8)(t^2).
To find the time it takes for the camera to reach the ground, you need to solve this equation for t. Rearrange the equation to isolate t and solve for it using algebraic techniques.
So, the correct equation to use for this problem is a) Y=Yo + Vo(t) + (1/2)(-g)(t^2).