forces help please im lost!?

Question

A person jumps from the roof of a house 4.5 m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 m. The mass of his torso (excluding legs) is 41 kg.
(a) Find his velocity just before his feet strike the ground.

(b) Find the average force exerted on his torso by his legs during deceleration.
Magnitude

direction?
_right
_left
_up
_down

Answer 1

V (before) = g t ; D = 1/2 g t**2 or V**2 = 2 g D
2.12
V = SQRT { 2(9.8) 4.5 } m/s = 9.39m/s

The momentum at this time is 41 kg ( 9.39m/s ) = 385 kg m/s

Deceleration = D ; V**2 = 2 D * 0.70m; D = (1/2) (88.2 m m /s s)(0.70 m ) : Deceleration D = 30.9 m/s**2

F * T = momentum = 385 kg m/s
T = time to decelerate to zero velocity
(1/2)D T**2 = 0.70 m; T = Sqrt { (2)(0.70m)/D}

F = 385 kg m /s / ([ Sqrt {2)(0.7m(/ 30.9 m = 1808 kg m / s

Answer 2

His velocity should be -9.8m/s^2 while the magnitude of the force can be solved using newton's laws. Gravity is constant.

I would use the law of vectors to find this. Draw a coordinate plane, with x and y components and the force vector will be the resultant. Set the resultant as your hypotenuse and solve for it. Set the angle of interest to be the one made by him and the ground with the horizontal.

Be sure to keep your values positive. I personally would put my data in the upper right quadrant ( 1 ) I think.

That is how I would solve it.

Answer 3

yeah I'm in the forces mate and I hate small print, so just get ya self a compass and you'll be fine!