Question

honors physics homework
conservation of energy
a classic example in physics is that a penny dropped from a tall building will fatally hit someone in
the head. let’s examine this using a penny (m = 0.0025 kg) dropped from the sears tower (h = 405
m). ignoring air resistance, you will find the potential, kinetic, and total energy as it falls to the
ground. in your calculations, use g = 10 m/s/s.

1. write a prediction of what would happen if a penny hit someone in the head if

dropped from a tall building. explain why.

1. complete the following table

to find the height: $h = 405 - \frac{1}{2}gt^2$
to find velocity: $v = gt$

time (s)	height (m)	velocity (m/s)	potential energy (j)	kinetic energy (j)	total energy (j)
2	385	20
4	325	40
6	225	60
9	0	90	0

Explanation:

Part 1: Prediction (Answer-Explanation Format)

A penny dropped from a tall building, if it hits someone's head, could cause injury (like a bruise, laceration, or even more severe trauma). This is because as the penny falls, gravitational potential energy ($PE = mgh$) converts to kinetic energy ($KE=\frac{1}{2}mv^{2}$). With height $h = 405\ m$, mass $m = 0.0025\ kg$, and $g = 10\ m/s^{2}$, the velocity at impact (using $v=gt$, and time to fall $t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times405}{10}}\approx9\ s$, so $v = 10\times9 = 90\ m/s$) is high. The kinetic energy at impact is $\frac{1}{2}\times0.0025\times90^{2}=101.25\ J$, which is enough to transfer significant force to the head.

Answer:

A penny dropped from a tall building hitting someone’s head could cause injury (e.g., bruising, laceration) because it gains high kinetic energy (from potential energy conversion) during the fall, transferring significant force at impact.

Part 2: Completing the Table (Step-by-Step Format)

We use the formulas:

• Height: $h = 405-\frac{1}{2}gt^{2}$
• Velocity: $v = gt$
• Potential Energy (PE): $PE=mgh$
• Kinetic Energy (KE): $KE=\frac{1}{2}mv^{2}$
• Total Energy (TE): $TE = PE + KE$ (or constant, equal to initial PE since energy is conserved)

Given $m = 0.0025\ kg$, $g = 10\ m/s^{2}$.

Step 1: Time $t = 0\ s$

• Height: $h = 405-\frac{1}{2}(10)(0)^{2}=405\ m$ (given)
• Velocity: $v = 10(0)=0\ m/s$ (given)
• PE: $PE=(0.0025)(10)(405)=10.125\ J$ (note: the handwritten "0" was incorrect; initial PE is non - zero)
• KE: $KE=\frac{1}{2}(0.0025)(0)^{2}=0\ J$
• TE: $TE = 10.125+0 = 10.125\ J$

Step 2: Time $t = 2\ s$

• Height: $h = 405-\frac{1}{2}(10)(2)^{2}=405 - 20=385\ m$ (given)
• Velocity: $v = 10(2)=20\ m/s$ (given)
• PE: $PE=(0.0025)(10)(385)=9.625\ J$
• KE: $KE=\frac{1}{2}(0.0025)(20)^{2}=0.5\ J$
• TE: $TE = 9.625 + 0.5=10.125\ J$

Step 3: Time $t = 4\ s$

• Height: $h = 405-\frac{1}{2}(10)(4)^{2}=405 - 80 = 325\ m$ (given)
• Velocity: $v = 10(4)=40\ m/s$ (given)
• PE: $PE=(0.0025)(10)(325)=8.125\ J$
• KE: $KE=\frac{1}{2}(0.0025)(40)^{2}=2\ J$
• TE: $TE = 8.125+2 = 10.125\ J$

Step 4: Time $t = 6\ s$

• Height: $h = 405-\frac{1}{2}(10)(6)^{2}=405 - 180 = 225\ m$ (given)
• Velocity: $v = 10(6)=60\ m/s$ (given)
• PE: $PE=(0.0025)(10)(225)=5.625\ J$
• KE: $KE=\frac{1}{2}(0.0025)(60)^{2}=4.5\ J$
• TE: $TE = 5.625+4.5 = 10.125\ J$

Step 5: Time $t = 9\ s$

• Height: $h = 405-\frac{1}{2}(10)(9)^{2}=405 - 405 = 0\ m$ (given)
• Velocity: $v = 10(9)=90\ m/s$ (given)
• PE: $PE=(0.0025)(10)(0)=0\ J$ (given)
• KE: $KE=\frac{1}{2}(0.0025)(90)^{2}=101.25\ J$
• TE: $TE = 0 + 101.25=101.25\ J$ (Wait, earlier TE was $10.125\ J$? No, mistake! Wait, initial PE at $t = 0$: $mgh=0.0025\times10\times405 = 10.125\ J$. At $t = 9\ s$, $h = 0$, so $PE = 0$, $KE=\frac{1}{2}\times0.0025\times90^{2}=101.25\ J$? That’s a contradiction. Wait, error in time calculation! Wait, $t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times405}{10}}=\sqrt{81}=9\ s$, correct. But $v = gt=90\ m/s$, correct. But $KE=\frac{1}{2}mv^{2}=\frac{1}{2}\times0.0025\times8100 = 10.125\ J$! Oh, I miscalculated $90^{2}=8100$, not $900$. So $KE=\frac{1}{2}\times0.0025\times8100 = 10.125\ J$. That matches initial PE. My bad earlier. So recalculating KE for $t = 9\ s$: $\frac{1}{2}\times0.0025\times90^{2}=\frac{1}{2}\times0.0025\times8100 = 10.125\ J$.

Correcting KE for all times:

• At $t = 2\ s$: $v = 20\ m/s$, $KE=\frac{1}{2}\times0.0025\times400 = 0.5\ J$ (correct)
• At $t = 4\ s$: $v = 40\ m/s$, $KE=\frac{1}{2}\times0.0025\times1600 = 2\ J$ (correct)
• At $t = 6\ s$: $v = 60\ m/s$, $KE=\frac{1}{2}\times0.0025\times3600 = 4.5\ J$ (correct)
• At $t = 9\ s$: $v = 90\ m/s$, $KE=\frac{1}{2}\times0.0025\times8100 = 10.125\ J$ (correct, matches initial PE)

So the table (filling in correct values, especially TE which is constant at $10.125\ J$ except at $t = 9\ s$? No, TE is always $PE + KE=10.125\ J$ (conservation of energy). Wait, at $t = 9\ s$, $PE = 0$, $KE = 10.125\ J$, so TE = $10.125\ J$. My earlier miscalculation of $90^{2}$ was wrong.

Now, the correct table entries:

Time (s)	Height (m)	Velocity (m/s)	Potential Energy (J)	Kinetic Energy (J)	Total Energy (J)
2	385	20	$0.0025\times10\times385 = 9.625$	$\frac{1}{2}\times0.0025\times20^{2}=0.5$	$10.125$
4	325	40	$0.0025\times10\times325 = 8.125$	$\frac{1}{2}\times0.0025\times40^{2}=2$	$10.125$
6	225	60	$0.0025\times10\times225 = 5.625$	$\frac{1}{2}\times0.0025\times60^{2}=4.5$	$10.125$
9	0	90	$0$	$\frac{1}{2}\times0.0025\times90^{2}=10.125$	$10.125$

Final Table (Filled)

Time (s)	Height (m)	Velocity (m/s)	Potential Energy (J)	Kinetic Energy (J)	Total Energy (J)
2	385	20	9.625	0.5	10.125
4	325	40	8.125	2	10.125
6	225	60	5.625	4.5	10.125
9	0	90	0	10.125	10.125