A man of mass 90 kg drops to a concrete patio from a window 0.78 m above the patio. He neglects to bend his knees on landing, taking 2.4 cm to stop. (a) What is his average acceleration from when his feet first touch the patio to when he stops? (b) What is the magnitude of the average stopping force exerted on him by the patio?
can u help me out? thanks lol
2007-03-15 22:05:28 · 4 answers · asked by Wilson J 4 in Science & Mathematics ➔ Physics
His speed when he touches the ground is
v = sqrt(2da) = sqrt(2*0.78*9.8) = 3.9m/s
His deceleration is
a = v^2/2d = 317 m/s^2
The average stopping force is
F = ma = 90kg * 317m/s^2 =28 530N
2007-03-16 02:09:23 · answer #1 · answered by catarthur 6 · 0⤊ 1⤋
mass of man = 90kg
distance from windows to patio = 0.78m
a)
mgh = 1/2mv^2
gh = 1/2 v^2
(9.81)(0.78) = 1/2 v^2
v^2 = 15.3036
speed of man when reach patio, v = 3.912 m/s
man take 2.4cm to stop.
v^2 = u^2 + 2as
0 = 15.3036 + 2a(2.4/100)
average accelaration, a = - 318.825 m/s^2
b)
magnitude of average stopping force,
F = ma + mg
F = m(a + g)
F = (90)(318.825 + 9.81)
F = 29.577 kN
2007-03-15 22:59:59 · answer #2 · answered by seah 7 · 0⤊ 0⤋
Use v^2 = u^2 + 2as to work out his speed as he starts to hit the ground.
Here u is the initial velocity, which is zero. a is the acceleration experienced which is due to gravity (a = g = 9.8 m/s^-2), and s is the distance travelled or 0.78m.
You should get v = 3.9 m/s
Then you need to work out the deceleration involved in going from a speed of 3.9 m/s to zero over a distance of 2.4 cm or 0.024 m. (Hint: use the same formula as before).
You should get -318.5 m/s^2
Finally, to calculate the force, use Newton's law F = ma
F = (90)(318.5) = 28,665 N.
This is a pretty large force to experience.
2007-03-15 22:55:04 · answer #3 · answered by dudara 4 · 0⤊ 0⤋
i think you need to contact CSI ... why does this man do that ... is it suoside or homocide ..???
2007-03-15 22:18:37 · answer #4 · answered by cdfxtech 3 · 0⤊ 2⤋