Question

A 75 lb (34 kg) boy falls out of a tree from a height of 10 ft (3 m). i. What is the kinetic energy of the boy when he hits the ground? Round your answer to the nearest joule. ii. What is the speed of the boy when he hits the ground? Round your answer to two significant figures. iii. Using the conversion factors of 1 m = 1.094 yd and 1 mi = 1760 yd, calculate the speed of the boy in miles per hour when he hits the ground.

Answer 1

Answer:

Kinetic energy of boy just before hitting the ground is $\approx$1000 J.

Speed of boy just before hitting the ground is 7.67 m/s

or 17.16 mi/hr.

Explanation:

Given that:

Mass of boy = 75lb = 34 kg

Height, h = 10ft = 3m

To find:

Kinetic energy of boy when he hits the ground.

As per law of conservation of energy The potential energy gets converted to kinetic energy.

$\therefore$ Kinetic energy at the time boy hits the ground = Initial potential energy of the boy when he was at the Height 'h'

The formula for potential energy is given as:

$PE = mgh$

Where m is the mass

g is the acceleration due to gravity, g = 9.8 $m/s^2$

h is the height of object

Putting all the values:

PE = $34 \times 9.8 \times 3 \approx 1000\ J$

Hence, Kinetic energy is $\approx$1000 J.

Formula for Kinetic energy is:

$KE = \dfrac{1}{2}mv^2$

where m is the mass and

v is the speed

Putting the values and finding v:

$1000 = \dfrac{1}{2}\times 34 \times v^2\\\Rightarrow v^2 = 58.82\\\Rightarrow v = 7.67\ m/s$

Given that:

1 m = 1.094 yd and 1 mi = 1760 yd

$\Rightarrow 1609\ m = 1\ mi$

Converting 7.67 m/s to miles/hour:

$\dfrac{7.67 \times 3600}{1609}=17.16\ mi/h$