JEE Main202228 Jul 2022Evening ShiftPhysicsMotion in One DimensionActual
A ball is thrown vertically upwards with a velocity of 19 . 6 m s - 1 from the top of a tower. The ball strikes the ground after 6 s . The height from the ground up to which the ball can rise will be k 5 m . The value of k is _____ (use g = 9 . 8 m s - 2 )
Correct answer
0
Step-by-step solution
Initial velocity of ball is u = 19 . 6   m   s - 1 , final velocity v = 0 . Using v = u + a t , here, acceleration a = - g . Time taken in upward motion above tower is t a = u g = 19 . 6 9 . 8 = 2   s Now, time taken from top most point to ground is t d = 6 - 2 = 4   s . Or, t d = 2 h max g (Using s = u t + 1 2 g t 2 ) ⇒ h max = 16 × 9 . 8 2 = 392 5   m . Hence, the value of k = 392 .