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A person throws balls into air vertically upward in regular intervals of time of one second. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is (Assume, g = 10 ms-2)

A person throws balls into air vertically upward in regular intervals of time of one second. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is (Assume, g = 10 ms-2) Correct Answer 5 m

CONCEPT:

  • When a body moves in a straight line under constant acceleration, it follows three equations of motion.

All unknowns are found by these equations.

  1. v = u + at;
  2. s = ut + (1/2) at2 
  3. v2 = u2 + 2as

where u is initial velocity, v is final velocity, a is constant acceleration, t is time and s is displacement.

CALCULATION:

  • The person throws balls into the air vertically upward at regular intervals of time of one second.
  • And the next ball is thrown when the velocity of the ball thrown earlier becomes zero.
  • So time to reach the highest point is 1 sec. If the initial velocity is u then at the highest point v will be zero,

So by 1st equation of motion:

0 = u - gt

u = gt = 10 × 1 = 10 m/s

The height to which the ball will rise at this initial speed (at the highest point v will be zero)

using the 3rd equation of motion:

02 = u2 - 2gh

102 = 2 × 10 × h

h = 5 m

So the correct answer is option 1.

[ alt="MISTAKE POINT" src="//storage.googleapis.com/tb-img/production/20/06/MISTAKE%20POINT.png" style="width: 151px; height: 45px;"> 

  • For all the equations of motion written in the concept part, keep in mind that the value of acceleration is constant. If acceleration is not constant, we can not use this equation.

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