Question

A skydiver jumps from an airplane at an altitude of 2,500 ft. He falls under the force of gravity until he opens his parachute at an altitude of 1,000 ft. Approximately how long does the jumper fall before he opens his chute

Answer 1

Answer: t = 9.68 s

Explanation:

The skydiver falls under force of gravity. Initial velocity = 0.

Displacement of the skydiver (when only under gravity),

s= 2500 ft - 1000 ft = 1500 ft

Acceleration due to gravity, a = 32 ft/s²

Using second equation of motion:

s = u t + 0.5 a t²

we can find out the time taken before the skydiver opens his parachute:

⇒1500 ft= 0 + 0.5 × 32 ft/s²×t²

⇒ t = 9.68 s

Answer 2

Answer:

t=9.7 seconds

Step-by-step explanation:

It is given that A skydiver falls under the force of gravity, therefore the initial velocity will be equal to zero that is:

$u=0$

Now,  Displacement of the skydiver (when only under gravity),

$s=2500-1000$

$s=1500ft$

Also, acceleration due to gravity is given as:

$a=32 fts^{-2}$

Using the second equation of motion, we have

$s=ut+\frac{1}{2}at^2$

Substituting the given values, we get

$1500=0+0.5(32)(t^2)$

⇒$1500=16t^2$

⇒$93.75=t^2$

⇒$t=9.7seconds$

Thus, approximately at the 9.7 seconds the jumper falls before he opens his parachute.