KinematicsEasy
Question
A ball is thrown vertically upwards from the foot of a tower. It crosses the top of the tower twice after an interval of 4s and reaches the foot of the tower 8s after it was thrown. What is the height of the tower ? Take g = 10 ms-2 :-
Options
A.60 m
B.80 m
C.100 m
D.120 m
Solution
Let h = AB be the height of the tower and P be the highest point reached (figure). The time taken by the ball to go from B to P = 4/2 = 2s and the time taken to go from A to P = 8/2 = 4s. Therefore, time taken by the ball to go from A to B is t = 4 - 2 = 2s.
If u is the velocity of projection, then
0 = u - 10 × 4 ⇒ u = 40 ms-1
∴ h = ut +
gt2
= 40 × 2 +
(-10) (2)2 = 60 m
If u is the velocity of projection, then
0 = u - 10 × 4 ⇒ u = 40 ms-1
∴ h = ut +
gt2= 40 × 2 +
(-10) (2)2 = 60 mCreate a free account to view solution
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