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a woman throws her keys straight up at 8.49 m/s. her friend catches them 2.89 m higher. how long were the keys in the air? (unit = s)​

Respuesta :

Nickiroo
Alright this one is just kinematics, you can use one of the big four equations and in this case this one works because acceleration is constant and you don’t have the final velocity to worry about:
Δx= displacement
Vi= initial velocity
T= time
A= acceleration

Δx=(Vi)(t)+(1/2)(a)(t^2)

A=9.8m/s2
Vi=8.49 m/s
Δx=2.89 meters

Then plug in:

2.89=(8.49)(t)+(1/2)(9.8)(t^2)

Then solve for t and you’re all set.

Answer:

0.465 seconds (Answer containing 3 significant figures)

Explanation:

Vi = 8.49

DeltaY = 2.89

A = -9.8

T = ?

Vf = Unknown

First Solve for Vf

Vf^2 = Vi^2 + 2aDeltaY

Vf = SQRT 8.49^2+2(-9.8)(2.89)

Vf = 3.93 m/s

Now find t

Vf  = Vi + at

3.93 = 8.49 + (-9.8)(t)

= 0.4653

= 0.465 s

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