Next confusing problem :\?

If Bob free falls for 2.6 seconds, what will be his final velocity and how far will he fall?

I don't even know where to start >.>

?2012-06-21T11:18:59Z

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Not sure but the equation must be z = (1/2)gt². The derivative of z is the velocity v = gt.

g is the gravity, and g = 9.86 m/s².

Hence v = 2.6 x 9.81 = 25.51m/s

Julian Javier2012-06-21T11:21:10Z

vf = vi - at

Assumming the initial velocity is 0, and gravity is 9.8 m/s^2

Vf = 0 + 9.8*2.6 = 25.48 m/s (because it's falling downwards)

Xf = Xi + vi*t - 1/2 g t^2

Asumming Xi = 0

Xf = 0 + 0 + 1/2(9.8)(2.6)^2 = 33.124 meters

??????2012-06-21T11:19:46Z

v = sqrt(2 g h) = sqrt(2 * 9.81 * h) (law of fall under gravity of Torricelli)

t = sqrt(2 h / g) = sqrt(2 * h / 9.81) = 2.6

=> h = 9.81 * 2.6² / 2 = 33.1578 m

=> v = sqrt(2 * 9.81 * 33.1578) = 25.506 m/s

So Bob will fall 33.1578 m in that time and reach a velocity of 25.506 m/s.

Anonymous2012-06-21T11:15:39Z

Gravity is a constant force of about 10 meters per second.

notthejake2012-06-21T11:18:33Z

a = -9.8 m / sec^2
v = -9.8 t + vo (vo = initial velocity, assume it to be 0)
h = -4.9t^2 + h0 (h0 = initial position, assume it to be 0)

v(2.6) = -9.8(2.6) = 25.48 m / sec
h(2.6) = -4.9(2.6)^2 = 19.6 m

(or, if you want it in feet, replace 9.8m / sec^2 with 32.2 ft / sec^2)

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