Question

An athlete leaps over a hurdle of height 1.0 m. The athlete leaps at a distance of 2.0 m before the hurdle and lands 2.0 m after the hurdle. If the angle of the leap is at 45 degree with the horizontal line, what is the speed at which the athlete leaps?

Answer 1

An athlete leaps over a hurdle of height $1.0\mathrm{m}$. The athlete leaps at a distance of $2.0\mathrm{m}$ before the hurdle and lands $2.0\mathrm{m}$ after the hurdle. If the angle of the leap is at 45 degree with the horizontal line, what is the speed at which the athlete leaps?

Solution:

$H = 1m -$ height of the hurdle

$L = 2m + 2m = 4m -$ distance of the leap

$\alpha = 45{}^{\circ} -$ angle of the leap with the horizontal line

The equation of motion along the X-axis:

$$x: L = V _ {x} t$$ $$V _ {x} = V \cdot \cos \alpha \text{ (from the right triangle $ABC$) } \rightarrow$$ $$x: L = V t \cos \alpha$$ $$t = \frac {L}{V \cos \alpha} (1)$$

The equation of motion along the Y-axis:

$$y: h = V _ {y} t - \frac {g t ^ {2}}{2}$$ $$V _ {y} = V \cdot \sin \alpha \text{ (from the right triangle $ABC$) } \rightarrow$$ $$h = V t \sin \alpha - \frac {g t ^ {2}}{2} (2)$$

(2)in (1):

$$h = V \frac {L}{V \cos \alpha} \sin \alpha - \frac {g \left(\frac {L}{V \cos \alpha}\right) ^ {2}}{2}, \text{which rearranges to give}$$ $$h = L \tan \alpha - \frac {g L ^ {2}}{2 V ^ {2} \cos^ {2} \alpha}$$ $$\frac {g L ^ {2}}{2 V ^ {2} \cos^ {2} \alpha} = L \tan \alpha - h$$ $$V ^ {2} = \frac {g L ^ {2}}{2 \cos^ {2} \alpha \cdot (L \tan \alpha - h)};$$ $$\cos 4 5 ^ {o} = \sin 4 5 ^ {o} = \frac {1}{\sqrt {2}}; \tan 4 5 ^ {o} = 1$$ $$V = \sqrt {\frac {g L ^ {2}}{2 \cos^ {2} \alpha \cdot (L \tan \alpha - h)}} = \sqrt {\frac {9 . 8 \frac {m}{s ^ {2}} \cdot (4 m) ^ {2}}{2 \cdot \frac {1}{2} \cdot (4 m - 1 m)}} = 7. 2 3 \frac {m}{s}$$

Answer: speed at which the athlete leaps is $V = 7.23\frac{m}{s}$