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Chapter 2: Q82GP (page 22)

You stand at the top of a cliff while your friend stands on the ground below you. You drop a ball from rest and see that she catches it 1.4 s later. Your friend then throws the ball up to you, such that it just comes to rest in your hand. What is the speed with which your friend threw the ball?

Short Answer

Expert verified

The obtained speed of the ball is \(u = 13.734\;{\rm{m/s}}\).

Step by step solution

01

Kinematic relation for determining the speed

In this problem, the upward direction is considered positive, and the origin is taken to be at the level where the ball was thrown.

Given data:

The time at which she catches the ball is\(t = 1.4\;{\rm{s}}\).

The relation from the kinematics equation is given by:

\(v = u + gt\)

Here, v is the speed of the ball whose value is zero because the ball is at rest, u is the required initial speed of the ball, and g is the gravitational acceleration.

02

Calculation of initial speed of the ball

On plugging the values in the above relation, you get:

\(\begin{aligned}0 = u + \left( { - 9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {1.4\;{\rm{s}}} \right)\\u = 13.734\;{\rm{m/s}}\end{aligned}\)

Thus, \(u = 13.734\;{\rm{m/s}}\) is the required speed.

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