Question

5. A sandbag is dropped from a hot air balloon that is 315 m above the ground. When the sandbag is dropped, the balloon is rising at 11 m/s. Find (a) the maximum height reached by the bag, (b) its position (relative to the ground) and velocity 4.3 s after it is released, and (c) when the bag hits the ground. PLEASE GO STEP BY STEP AND WRITE THE EQAUTION SO I UNDERSTAND HOW YOU GOT TO THE SOULTION! :) I AM TRYING TO LEARN THE INFO!

Answer 1

The sandbag has the same initial velocity as that of the balloon.

(a) At the maximm height, the sandbag turns around from moving up to moving down. Momentarily at the maximum height the velocity of the sandbag is zero.

The sandbag is under uniform gravitational acceleration of the earth . The motion of the sandbag is governed by the Kinematics equations for uniformly accelerated motion. We use

Solving for

(b) The velocity of the sandbag after time t with the help of initial velocity can be determined by the following kinematics equation.

The position of the sandbag after time t is given by the following kinematics equation

(c) The position of the sandbag when it hits the ground is .The velocity of the sandbag at this point can be calculated by the following kinematics equation