A ball thrown up vertically returns to the thrower after 12 second. Find
(i) velocity with which it was thrown up.
(ii) the maximum height it reaches.
(iii) its position after 4 s
(Take g = 10 m s–2)
Step 1: Given data:
Time for the ball to return is 12 seconds.
Therefore,
⸫ Time taken to reach the maximum height =
=
Since it takes for the ball from the ground to maximum height. Then,
Step 2: (i) Finding Initial Velocity
Let Initial velocity = u
At the highest point of motion, velocity becomes 0
Therefore
Final velocity = v = 0
Acceleration = (Acceleration is negative because ball is thrown upwards)
Time =
We know v, a and t
Finding u by 1st equation of motion
Thus, the initial velocity is
Step 3: (ii) Calculating the maximum height
Calculating maximum height
We know v, a and t
Finding distance using 2nd equation of motion
s =
=
=
=
s = 180 m
Thus, the maximum height it reaches is 180m.
Step 4: (iii) Find the position after 4 s.
The ball reaches the highest point in 6 seconds.
At 4 seconds, it will be moving upward.
Initial velocity = 60
Time = 4 seconds
Acceleration =
We need to find s
We know u, a and t
Finding using 2nd equation of motion.
Thus, the position of the ball after 4 seconds is 160 m.