Question

A 0.14-kg baseball is dropped from rest from a height of 22 m above the ground. What is the magnitude of its momentum just before it hits the ground if we neglect air resistance? Use g = 9.8 m/s2

Answer 1

Approximately
`2.9 \; {\rm kg\cdot m\cdot s^(-1)}`.

Step-by-step explanation:

Under the assumptions, the baseball will accelerate downwards at a constant acceleration
`a` of
`a = g = 9.8\; {\rm m\cdot s^(-2)}`.

Let
`x` denote the displacement of the baseball. Let
`u` denote the initial velocity of the baseball. Let
`v` denote the velocity of the baseball right before landing.

It is given that the baseball in this question was initially at rest, such that
`u = 0\; {\rm m\cdot s^(-1)}`. It is also known that the displacement was
`x = 22\; {\rm m}`. The SUVAT equation
`v^(2) - u^(2) = 2\, a\, x` relates these quantities.

Rearrange this equation to find the velocity of the baseball right before landing:

`\begin{aligned}v &= \sqrt{2\, a\, x + u^(2)} \\ &= \sqrt{2 \, (9.8\; {\rm m\cdot s^(-2)})\, (22\; {\rm m}) + (0\; {\rm m\cdot s^(-1)})^(2)} \\ &\approx 20.77\; {\rm m\cdot s^(-1)}\end{aligned}`.

When an object of mass
`m` travels with a velocity of
`v`, the momentum
`p` of that object will be
`p = m\, v`.

Thus, the momentum of this baseball right before landing will be:

`\begin{aligned}p &= m\, v \\ &\approx (0.14\; {\rm kg})\, (20.77\; {\rm m\cdot s^(-1)}) \\ &\approx 2.9\; {\rm kg\cdot m\cdot s^(-1)}\end{aligned}`.