Basic Maths and Units and DimensionsHard
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
View Solution FreeTopic: Basic Maths and Units and Dimensions·Practice all Basic Maths and Units and Dimensions questions
More Basic Maths and Units and Dimensions Questions
Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere...A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitudeand length. ...If a unit vector is represented by then the value of ′α′ is :-...Two sitar strings A and B playing the note ′Ga′ are slightly out of tune and produce beats of frequency 6 Hz...You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face and views the magnified ima...