Question

A) If the camera is 44 m above the ground when it is dropped, how long does it take for the camera to reach the ground?
B) If the camera is 44 m above the ground when it is dropped, what is its velocity just before it lands? Let upward be the positive direction for this problem.

Answer 1

Let's describe all the givens first. All the givens are about the camera.

initial velocity (v_i) = -1.5m/s, since the climber is climbing DOWNWARDS.

initial position (x_i)= 44m, since the climber drops the camera at altitude 44 m.

final position (x_f) = 0m, since the camera hits the ground.

acceleration (a) = -9.81 m/s², since the acceleration of gravityis -9.81 m/s.

time (t) = ?

final velocity (v_f) = ?

You are trying to find time and the final velocity.

a. Use the equation x_f= x_i+ v_iT + 0.5aT²

0 = 44 + (-1.5)(T) + (0.5)(-9.81)(T²)

-4.905T²- 1.5T + 44 = 0

If you solve using the quadratic formula, you get T = -3.15 or 2.85

Since time can't be negative, the answer is 2.85 seconds.

b. Use the equation v_f= v_i+ aT

v_f= -1.5 + (-9.81)(2.85)

v_f= -29.46 m/s.

Answer 2

a)4.4898s