Question

On a distant planet, where the gravitational acceleration has magnitude 10.0 m/s², a basketball center holds a basketball straight out, 1.9m above the floor, and releases it. It bounces off the floor and rises to a height of 1.1m. What is the ball's velocity just before it hits the floor? (Let up be the positive direction throughout this problem.) answer in m/s ( ± 0.1 m/s)

Answer 1

Approximately
`6.2\; {\rm m\cdot s^(-1)}` downward (assuming air resistance is negligible; rounded to one decimal place as the question requested.)

Step-by-step explanation:

Under the assumptions, the ball would be accelerating downward at
`a = (-g) = (-10.0)\; {\rm m\cdot s^(-2)}`, where
`g = 10\; {\rm m\cdot s^(-1)}` is the strength of the gravitational field.

Apply the SUVAT equation
`v^(2) - u^(2) = 2\, a\, x` to find the velocity just before landing. In this equation:

• 
  `v` is the final velocity,
• 
  `u` is the initial velocity,
• 
  `a` is acceleration, and
• 
  `x` is displacement (change in position.)

This question implies that the initial velocity was
`u = 0\; {\rm m\cdot s^(-1)}` by stating the basketball was released from the initial position.

The basketball was initially at a height of
`1.9\; {\rm m}`. RIght before hitting the ground, the height would be
`0\; {\rm m}`. Subtract the initial position from the new one to find displacement (change in position) of the basketball:
`x = 0\; {\rm m} - 1.9\; {\rm m} = (-1.9)\; {\rm m}`. The value of displacement
`x` is negative because the new position of the basketball is below where it was initially released.

Rearrange the equation
`v^(2) - u^(2) = 2\, a\, x` to find an expression for
`v`, the velocity of the basketball right before hitting the ground:

`\displaystyle v^(2) = u^(2) + 2\, a\, x`.

Note that there are two possible roots for this equation: a negative one (pointing downward) and a positive one (pointing upward.) However, only the negative one is valid because the basketball would be travelling downward before hitting the ground:

`\displaystyle v = -\sqrt{u^(2) + 2\, a\, x}`.

Substitute in
`u = 0\; {\rm m\cdot s^(-1)}`,
`a = (-10)\; {\rm m\cdot s^(-1)}`, and
`x = (-1.9)\; {\rm m}` into this expression. Evaluate to find the value of
`v`:

`\begin{aligned} v &= -\sqrt{u^(2) + 2\, a\, x} \\ &= -\sqrt{0^(2) + 2\, (-10)\, (-1.9)}\; {\rm m\cdot s^(-1)} \\ &\approx (-6.2)\; {\rm m\cdot s^(-1)}\end{aligned}`.

In other words, the velocity of the basketball would be approximately
`6.2\; {\rm m\cdot s^(-1)}` downward toward right before hitting the ground.