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

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Answer:
The velocity of the skateboard is 0.774 m/s.
Explanation:
Given that,
The spring constant of the spring, k = 3086 N/m
The spring is stretched 18 cm or 0.18 m
Mass of the student, m = 100 kg
Potential energy of the spring, 
To find,
The velocity of the car.
Solution,
It is a case of conservation of energy. The total energy of the system remains conserved. So,






v = 0.774 m/s
So, the velocity of the skateboard is 0.774 m/s.
Answer:
250 N
433 N
Explanation:
N = Normal force by the surface of the inclined plane
W = Weight of the block = 500 N
f = static frictional force acting on the block
Parallel to incline, force equation is given as
f = W Sin30
f = (500) Sin30
f = 250 N
Perpendicular to incline force equation is given
N = W Cos30
N = (500) Cos30
N = 433 N
Answer:
300 m
Explanation:
The train accelerate from the rest so u = 0 m/sec
Final speed that is v = 80 m/sec
Time t = 30 sec
The distance traveled by first plane = 1200 m
We know the equation of motion
where s is distance a is acceleration and u is initial velocity
Using this equation for first plane 

As the acceleration is same for both the plane so a for second plane will be 2.67 
The another equation of motion is
using this equation for second plane 
s = 300 m