Vertical impulse force in triple jump takeoff

[JJBladester]

Homework Statement

The triple jump is a track-and-field event in which an athlete gets a running start and tries to leap as far as he can with a hop, step, and jump. Shown in the figure is the initial hop of the athlete. Assuming that he approaches the takeoff line from the left with a horizontal velocity of 10 m/s, remains in contact with the ground for 0.18s, and takes off at a 50° angle with a velocity of 12 m/s, determine the vertical component of the average impulsive force exerted by the ground on his foot. Give your answers in terms of the weight W of the athlete.

Answer:
6.21W

Homework Equations

$$mv_1+\sum Imp_{1\rightarrow 2}=mv_2$$

The Attempt at a Solution

$$mv_1+\sum Imp_{1\rightarrow 2}=mv_2$$

$$mv_{1_y}+F_y\Delta t=mv_{2_y}$$

$$0+F_y(0.18)=m(12sin(50))$$

$$F_y=\frac{m(12sin(50))}{0.18}=51.1m$$

If all of the m's were replaced with W's (an equivalent way to write the equation above), you would end up with 51.1W. The book's answer was 6.21W.

 

[Delphi51]

Since W = mg, you should replace your m with W/g to get F = 5.21W.
I get the same value using the impulse equation
FΔt = mΔv
F*.18 = W/g*12*sin(50)

Looks like we are short by one W!
Ah, we have forgotten the extra W needed to just hold the guy up, or to cancel the force of gravity pulling him down.

 

[JJBladester]

Looks like we are short by one W!
Ah, we have forgotten the extra W needed to just hold the guy up, or to cancel the force of gravity pulling him down.

$$\sum F_y=m\frac{dv}{dt}$$

$$N-W=m\frac{dv}{dt}$$

$$Ndt-Wdt=\frac{W}{g}dv$$

$$N(t)-W(t)=\frac{W}{g}\left (v_2-v_1 \right )$$

$$N=W+\frac{\frac{W}{g}\left (v_2-v_1 \right )}{t}$$

$$N=W\left (1+\frac{\frac{(12sin50)}{9.81}}{0.18} \right )=6.21W$$

Ahhhh I can go on with my weekend now. You're a lifesaver Delphi51.

 

[Delphi51]

Most welcome.