Question:

A ball is thrown vertically upwards with a velocity of \(19.6 ms^{–1}\) 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 \(g = 9.8 m/s^2\))

Updated On: Jun 24, 2025
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Solution and Explanation

\(v = 19.6 m/s \)
\(t = 6s \)
Time taken in upward motion above tower = \(2s\) 
⇒ Time taken from top most point to ground = \(4s\)
⇒ \(\sqrt{\frac{2h}{g}}=4\)
\(h=\frac{16×9.8}{2}=8×9.8\)
⇒ \(k=8×9.8×5=392\)
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Concepts Used:

Motion in a straight line

The motion in a straight line is an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. It is nothing but linear motion. 

Types of Linear Motion:

Linear motion is also known as the Rectilinear Motion which are of two types:

  1. Uniform linear motion with constant velocity or zero acceleration: If a body travels in a straight line by covering an equal amount of distance in an equal interval of time then it is said to have uniform motion.
  2. Non-Uniform linear motion with variable velocity or non-zero acceleration: Not like the uniform acceleration, the body is said to have a non-uniform motion when the velocity of a body changes by unequal amounts in equal intervals of time. The rate of change of its velocity changes at different points of time during its movement.