A small steel ball is shot vertically upwards from the top of a building 25m above the ground with an initial velocity 18m/s
In what time it will reach the maximum height?(Ans t=1.83sec)
How high above the building will the ball rise?(Ans h=16.51m)
Compute the velocity with which it will strike the ground? And the total time it is in motion. (Ans v=28.538m/s,t2=2.91sec.t=4.74sec)
Answers
Answer:
Answers are approximately right in my solution. Just take g=9.8 instead of 10 then you will get the stated answers
Explanation:
Solution is attached to the answer. Please check it out. Tell me if it helped you
Answer:
The time taken by the ball to reach the maximum height is 1.83s.
The height above the building the ball will go to is 16.51m.
The velocity with which it will strike the ground is 28.538m/s.
The total time of the motion is 4.74s.
Explanation:
Given:
U = 18m/sec
S = 25m
Here,
The height above the building will the ball rise is denoted by H.
The height of the building is denoted by S.
The velocity at the height of H above the building is denoted by .
The time taken by the ball to reach the height of H above the building is denoted by t.
The Velocity at the ground is denoted by V.
The total time of the motion is denoted by T.
The acceleration due to gravity is denoted by g.
The initial speed is denoted by U.
Then,
For the time taken by the ball to reach the maximum height,
By the equation,
Then,
For the height above the building will the ball goes,
By the equation,
Then,
For the velocity with which it will strike the ground,
By the equation,
Then,
For the total time of the motion,
By the equation,
So,
The time taken by the ball to reach the maximum height is 1.83s.
The height above the building will the ball goes is 16.51m.
The velocity with which it will strike the ground is 28.538m/s.
The total time of the motion is 4.74s.