Question

An athlete jumps at an angle 30° with a maximum speed of 9.4m/s ✨a)what is the shape of the path followed by athlete in jumping? ✨b) Obtain an expression to calculate the horizontal range covered by the athlete. ✨c)Find the range covered by him in the above, suggest the angle by jump which the athlete can attain maximum range. ​

Answer 1

Athlete jumped with 9.4 m/s at angle 30° with horizontal.

• Now, shape of trajectory is PROJECTILE TRAJECTORY (shape: Parabola).

Now, for range :

$\sf \: y = u_{y}(t) - \dfrac{1}{2} g {t}^{2}$

$\sf \implies\: 0= u_{y}(t) - \dfrac{1}{2} g {t}^{2}$

$\sf \implies\: 0= u \sin( \theta) - \dfrac{1}{2} g t$

$\sf \implies\: u \sin( \theta) = \dfrac{1}{2} g t$

$\sf \implies\:t = \dfrac{2u \sin( \theta) }{g}$

Now, range will be :

$\sf \: R = u_{x}(t)$

$\sf \implies\:R = u \cos( \theta) \times \dfrac{2u \sin( \theta) }{g}$

$\sf \implies\:R = \dfrac{2 {u}^{2} \sin( \theta) \cos( \theta) }{g}$

$\boxed{ \sf \implies\:R = \dfrac{{u}^{2} \sin( 2\theta) }{g} }$

• This is the expression for RANGE.

Now, max range will be when $\theta$ = 45°.

$\sf \implies\:R_{max} = \dfrac{{u}^{2} }{g}$

$\sf \implies\:R_{max} = \dfrac{{(9.4)}^{2} }{9.8}$

$\sf \implies\:R_{max} = 9.01 \: m$

So, max range is 9.01 metres.