Question

a 10g ball is thrown directly downwards from a high bridge with an initial kinetic energy of 0.2 j. how long will it take to fall 20m

Answer 1

Approximately
`0.74\; {\rm s}` (assuming that air resistance is negligible and that
`g = 9.81\; {\rm m\cdot s^(-2)}`.)

Step-by-step explanation:

Apply the following steps to find the duration of this free fall:

• Apply unit conversion and ensure that all quantities are measured in standard units.
• Find the initial velocity of this ball from its initial kinetic energy.
• Apply the SUVAT equations to find the velocity of the ball after traversing the given distance.
• Divide the change in velocity by acceleration to find the time required.

The standard unit for mass is kilogram. Apply unit conversion and ensure that the mass of this ball is measured in standard units:

`\begin{aligned} m &= 10\; {\rm g} \\ &= 10\; {\rm g} * \frac{1\; {\rm kg}}{10^(3)\; {\rm g}} = 0.010 \; {\rm kg}\end{aligned}`.

When an object of mass
`m` travels at a speed of
`v`, the kinetic energy of that object would be
`(\text{KE}) = (1/2)\, m\, v^(2)`. Rearrange this equation to find speed in terms of kinetic energy and mass:

`\displaystyle v = \sqrt{\frac{2\, (\text{KE})}{m}}`.

In this question, the initial kinetic energy of this ball is
`0.2\; {\rm J}`. The mass of this ball is
`m = 0.010\; {\rm kg}`. Hence, the initial velocity of this ball would be:

`\begin{aligned}u &= \sqrt{\frac{2\, (\text{KE})}{m}} \\ &= \sqrt{\frac{2\, (0.2\; {\rm J})}{(0.010\; {\rm kg})}} \\ &= √(40)\; {\rm m\cdot s^(-1)} \\ &\approx 6.3246\; {\rm m\cdot s^(-1)}\end{aligned}`.

During the free fall, the ball accelerates at a constant
`a = g = 9.81\; {\rm m\cdot s^(-2)}` (downwards.) The following SUVAT equation relates the velocity before and after accelerating to displacement:

`v^(2) - u^(2) = 2\, a\, x`,

Where:

• 
  `v` is velocity after accelerating,
• 
  `u` is velocity before accelerating,
• 
  `a` is acceleration, and
• 
  `x` is displacement.

Rearrange this equation to find
`v`, velocity of the object after accelerating:

`\begin{aligned} v &= \sqrt{u^(2) + 2\, a\, x} \\ &= \sqrt{\left(√(40)\right)^(2) + 2\,, (9.81)\, (20)}\; {\rm m\cdot s^(-1)} \\ &\approx 20.79423\; {\rm m \cdot s^(-1)}\end{aligned}`.

Acceleration is the rate of change in velocity. To find the duration of the motion, divide the change in velocity by acceleration:

`\begin{aligned}t &= (v - u)/(2\, a) \\ &\approx (20.79423 - 6.3246)/(2\, (9.81))\; {\rm s} \\ &\approx 0.74\; {\rm s} \end{aligned}`.