A parachutist descending at a speed of 10.0 m/s loses a shoe at an altitude of 50.0 m.
a. When does the shoe reach the ground?
b. What is the velocity of the shoe just before it hits the ground?

Respuesta :

Answer:

The answer to your question is t = 2.33 s; vf = 32.87 m/s

Explanation:

Data

vo = 10 m/s

h = 50 m

t = ?

vf = ?

g = 9.81 m/s²

Formula

h = [tex]\frac{vo + vf}{2} t[/tex]

h = vot + [tex]\frac{1}{2} at^{2}[/tex]

a) Substitution

             50 = 10t + 1/2(9.81)t²

Simplification

             4.91t² + 10t - 50 = 0

Solve the equation using an online calculator

     t₁ = -4.33                t₂ = 2.33 s      

The correct answer is t₂, t₁ is not correct because there are not negative times.

       

b) vf² = vo² + 2gh

   vf² = 10² + 2(9.81)(50)

   vf² = 100 + 981

   vf² = 1081

   vf = √1081

   vf = 32.87 m/s