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
Answer:
True.
Explanation:
If the sum of the external forces on an object is zero, then the sum of the external torques on it must also be zero.
The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.
Hence, the given statement is true.
You want to draw a free body diagram of the forces on the sled in the horizontal x-direction.
If you visualize the system in an x-y coordinate plane, the force along the x-direction is the angle it makes with the x-axis multiples by the force.
The angle made with the x-axis is cosine of the angle theta.
Please see picture attached.
a)You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.
from above statement we got
height = 78.4 m
since the ball is thrown, so its vertical velocity would be zero
u = 0
taking g = 9.8m/s^2
now, using the equation of motion
h = ut + gt^2/2
now putting all the values in it
we got ,
78.4 = 9.8 * t^2/ 2
by solving we got,
t = 4 sec
b) now, since along the horizontal , no force acting and accelaration is zero so
R = ut , R is RANGE
R = 5 * 4
range = 20 m
c) vertical components of the stone’s velocity just before it hits the ground = v sin θ =
horizontal components of the stone’s velocity just before it hits the ground = v cos θ
To know more about velocity visit :
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