# Thread: physics motion homework help

1. ## physics motion homework help

A photographer in a helicopter ascending vertically at a constant rate of 12.5 m/s accidentally drops a camera out the window when the helicopter is 60 m above the ground. (a) How long will it take the camera to reach the ground? (b) What will it's speed be when it hits?

2. Hello, twistedmexican!

A photographer in a helicopter ascending vertically at a constant rate of 12.5 m/s
accidentally drops a camera out the window when the helicopter is 60 m above the ground.
(a) How long will it take the camera to reach the ground?
(b) What will it's speed be when it hits?
You're expected to know the "free fall" formula: . $y \;=\;h_o + v_ot - 4.9t^2$

. . where: . $\begin{Bmatrix}h_o & = & \text{initial height} \\ v_o & = & \text{initial velocity} \\ y & = & \text{height of object}\end{Bmatrix}$

We have: . $h_o = 60,\;v_o = 12.5$

. . Hence: . $y \;=\;60 + 12.5t - 4.9t^2$

(a) When the camera reaches the ground, its height is 0.

So we have: . $60 + 12.5t - 4.9t^2 \:=\:0$
. . a quadratic equation: . $4.9t^2 - 12.5t - 60 \:=\:0$
. . which factors: . $(t - 5)(4.9t + 12) \:=\:0$
. . and has the positive root: . $t \,=\,5$

It takes 5 seconds for the camera to crash to the ground.

(b) The velocity is given by the derivative: . $v \:=\:\frac{dy}{dt} \:=\:12.5 - 9.8t$
At $t = 5\!:\;v \:=\:12.5 - 9.8(5) \:=\:-36.5$

The speed at impact is 36.5 m/s . . . downward, of course.