Answer:
Velocity.
Explanation:
Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.
As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:
Horizontal range: As per expression:
R= (
*sin2θ)/g
the range depending on the square of the initial velocity.
Maximum height: As per expression:
H= ( * 
θ
)/2g
the maximum distance also depends upon square of the initial velocity.
Answer:
If I understand correctly. Line B is parallel to the circle. Also, the angle is less than 90.
- The size of the circle determines.
- The diameter should not be fixed either.
Answer:
2 kg
Explanation:
Remember:
F = m * a re-arrange to
F/a = m substitute in the given values
10 / 5 = 2 kg
The object does not move.
Given :
Walk in forward direction is 30 m .
Walk in backward direction is 25 m .
To Find :
The distance and displacement .
Solution :
We know , distance is total distance covered and displacement is distance between final and initial position .
So , distance travelled is :
D = 30 + 25 m = 55 m .
Now , we first move 30 m in forward direction and then 25 m in backward direction .
So , displacement is :
D = 30 - 25 m = 5 m .
Therefore , distance and displacement covered is 55 m and 5 m respectively .
Hence , this is the required solution .
