Question

An athlete jump an angle of ,
30° maximum speed is 9.4 m/ s. Find the horizontal range

Answer 1

Given :

• Angle at which athelete jumped = 30°
• Maximum speed = 9.4 m/s

To find :

Horizontal range of athelete

Formula used :

$\sf Range = \dfrac{u^2 \sin(2\theta)}{g}$

Here,

• u = speed
• g = acceleration due to gravity
• Theta = angle

Solution :

Putting value in above formula

$\sf \implies Range = \dfrac{(9.4 ms^{-1})^2 \times \sin(2 \times 30^\circ)}{10 \: ms^{-2}}$

$\sf \implies Range = \dfrac{(9.4 \times 9.4 \: {m}^{2} s^{-2}) \times \sin(60^\circ)}{10 \: ms^{-2}}$

$\sf \implies Range = \dfrac{(88.36\: {m}^{2} s^{-2}) \times \: \dfrac{ \sqrt{3} }{2} }{10 \: ms^{-2}}$

$\sf \implies Range = \dfrac{(88.36\: {m}^{2} s^{-2}) \times \dfrac{1.732}{2} }{10 \: ms^{-2}}$

$\sf \implies Range = \dfrac{(88.36\: {m} ) \times 0.866 }{10 }$

$\sf \implies Range = \dfrac{76.51\: {m} }{10 }$

$\sf \implies Range = \: 7.651 \: m$

ANSWER : 7.651 m