Answer:
Explanation:
(a) The velocity of object is zero when it is at maximum height.
(b) The direction of velocity changes as it starts moving downwards after it reaches the maximum height.
(c) Acceleration due to gravity always acts downwards so its sign remains same.
Mgh= 1/2 m v^2
gh= 1/2 v^2 9.8 * 8= 1/2 v^2
solve for v
V = sqrt(2gH)where H = 8 m.
Answer: now take this with a grain of salt because I'm in middle school but I think that the more massive object has more potential energy.
Explanation:
1. Velocity at which the packet reaches the ground: 121.2 m/s
The motion of the packet is a uniformly accelerated motion, with constant acceleration 
directed downward, initial vertical position 
, and initial vertical velocity 
. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:
substituting, we find
2. height at which the packet has half this velocity: 562.6 m
We need to find the heigth at which the packet has a velocity of
In order to do that, we use again the same SUVAT equation substituting 
with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:
Which means that the heigth of the packet was
