Question

question 1
a mans gravitational potential energy is 1870 j as he sits 2.20 m above the ground in a sky diving airplane before it takes off. what is his gravitational potential energy when be begins to jump from the airplane at an altitude of 923 m?
1.87×10³ j
3.29×10⁴ j
4.22×10⁶ j
7.85×10⁵ j
9.36×10² j

Explanation:

Step1: Find the man's mass

The formula for gravitational potential energy is \(U = mgh\). When \(U_1=1870\space J\) and \(h_1 = 2.20\space m\), we can find \(m\) from \(m=\frac{U_1}{gh_1}\). Taking \(g = 9.8\space m/s^{2}\), then \(m=\frac{1870}{9.8\times2.20}\)

$$m=\frac{1870}{21.56}\approx86.74\space kg$$

Step2: Calculate the new potential energy

Now, when \(h_2 = 923\space m\), use the formula \(U_2=mgh_2\). Substitute \(m = 86.74\space kg\) and \(g = 9.8\space m/s^{2}\) and \(h_2=923\space m\)

$$U_2=86.74\times9.8\times923$$

$$U_2=86.74\times9045.4$$

$$U_2\approx7.85\times 10^{5}\space J$$

Answer:

\(7.85\times 10^{5}\space J\) (the fourth option)