Question

Find the fall time for an object dropped from an altitude of 25,000 meters, neglecting air

drag (i.e., the time it takes the bullet in the previous · example to return to the starting

point, from the time it reached its maximum height).​

Answer 1

$71.42 \: s$

Explanation:

$Given, \: \: \: h = 25000 \: m, \: \: u = 0 \: m {s}^{ - 1} \\ and \: \: \: g = 9.8 \: m {s}^{ - 2} \\ we \: \: know \: \: \boxed{h = ut + \frac{1}{2}g {t}^{2} } \\ Now, \: \: \: putting \: \: values \\ 25000 = 0 \times t + \frac{1}{2} \times 9.8\times {t}^{2} \\ 25000 = 4.9 \times {t}^{2} \\ {t}^{2} = \frac{25000}{4.9} \\ \: \: {t}^{2} = \frac{250000}{49} \\ \: \: \: \: \: \: \: t = \sqrt{ \frac{250000}{49} } \\ \: \: \: \: \: \: = \frac{500}{7} \\ \: \: \: \: \: \: \: \: \: \: \: \: \:\: = \boxed{71.42 \: s}$