Question

free fall calculations:
free-fall equations:

1. ( h = v_i t + \frac { 1 } { 2 } g t ^ { 2 } )
2. ( v _ { f } ^ { 2 } = v _ { i } ^ { 2 } + 2 g h )
3. ( v _ { f } = v _ { i } + g t )
4. ( g = 9.8 \frac { m } { s ^ { 2 } } ) (on earth)
5. the observation deck of a skyscraper is 420 m above the street. determine the time required for a penny to free-fall from the deck to the street below.
6. dionte is riding the giant drop at great america. if dionte free-falls for 2.6 seconds: a. what will be his final velocity? b. how far will he fall?
7. on the moon, a feather is dropped from a height of 1.40 m. the acceleration due to gravity on the moon is ( 1.67 mathrm { m } / mathrm { s } ^ { 2 } ). determine the time it takes for the feather to fall to the surface of the moon.
8. a robot probe drops a camera off the rim of a 239 m high cliff on mars, where the free-fall acceleration is ( 3.7 mathrm { m } / mathrm { s } ^ { 2 } ). a. find the time required for the camera to reach the ground. b. find the velocity with which it hits the ground.

Explanation:

Problem 1

Step1: Identify the formula

Since the penny is in free - fall and we assume \(v_i = 0\) (starts from rest), we use the formula \(h=v_it+\frac{1}{2}gt^{2}\). Substituting \(v_i = 0\), the formula simplifies to \(h=\frac{1}{2}gt^{2}\).

Step2: Solve for \(t\)

Given \(h = 420m\) and \(g=9.8m/s^{2}\), from \(h=\frac{1}{2}gt^{2}\), we can solve for \(t\). First, rewrite the formula as \(t^{2}=\frac{2h}{g}\). Then \(t=\sqrt{\frac{2h}{g}}\).
Substitute \(h = 420m\) and \(g = 9.8m/s^{2}\) into the formula: \(t=\sqrt{\frac{2\times420}{9.8}}=\sqrt{\frac{840}{9.8}}\approx\sqrt{85.71}\approx9.26s\)

Step1: Identify the formula

We use the formula \(v_f=v_i + gt\). Since the initial velocity \(v_i = 0\) (starts from rest in free - fall), the formula simplifies to \(v_f=gt\)

Step2: Calculate \(v_f\)

Given \(g = 9.8m/s^{2}\) and \(t = 2.6s\), substitute into \(v_f=gt\). Then \(v_f=9.8\times2.6=25.48m/s\)

Step1: Identify the formula

We use the formula \(h=v_it+\frac{1}{2}gt^{2}\). Since \(v_i = 0\) (starts from rest), the formula simplifies to \(h=\frac{1}{2}gt^{2}\)

Step2: Calculate \(h\)

Given \(g = 9.8m/s^{2}\) and \(t = 2.6s\), substitute into \(h=\frac{1}{2}gt^{2}\). Then \(h=\frac{1}{2}\times9.8\times2.6^{2}=4.9\times6.76 = 33.124m\)

Answer:

The time \(t\) is approximately \(9.26s\)

Problem 2a