Answer:
44.3 m/s
Explanation:
Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.
What is the magnitude of its velocity just before it strikes the ground ?
The parameters given are:
Height H = 100m
Since the ball is thrown from a top of a building, initial velocity U = 0
Let g = 9.8m/s^2
Using third equation of motion
V^2 = U^2 + 2gH
Substitute all the parameters into the formula
V^2 = 2 × 9.8 × 100
V^2 = 200 × 9.8
V^2 = 1960
V = 44.27 m/s
Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately
On <span>Contact Forces: </span>
<span>Frictional Force </span>
<span>Tension Force </span>
<span>Normal Force </span>
<span>Air Resistance Force </span>
<span>Applied Force </span>
<span>Spring Force </span>
<span>Action-at-a-Distance Forces: </span>
<span>Gravitational Force </span>
<span>Electrical Force </span>
<span>Magnetic Force</span>
Answer : 413.44N
Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .
- From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
- When a elevator accelerates down , the weight recorded is less than the actual weight .
From the Free body diagram ,
- Mass of the man = 64.2 kg
