Question

A ball is thrown from a field with a speed of 12.0m/s at an angle of 45° with the horizontal. At what distance will it hit the field again? Take g = 10.0 m/s^2​

Answer 1

$\underline{\rm{\red{Question - }}}$

A ball is thrown a from a field with a speed of 12 m/s at an angle of 45° with the horizontal. At what distance will it hit the field again ? ( g = 10 m/s² )

$\underline{\rm{\red{Answer - }}}$

$\underline{\rm{\pink{Given - }}}$

• $\rm u = 12 m/s$
• $\rm \theta = 45\degree$
• $\rm g = 10 m/s^2$

where

u is initial velocity.

$\rm \theta$ is angle of projection.

g is acceleration due to gravity.

$\underline{\rm{\pink{To\: find - }}}$

• Horizontal range $\longrightarrow$ R

$\underline{\rm{\pink{Formula\: used -}}}$

$\boxed{\rm\red{R = \frac{u^2sin2\theta}{g}}}$

$\underline{\rm{\pink{Solution - }}}$

• $\rm u = 12 m/s$
• $\rm \theta = 45\degree$
• $\rm g = 10 m/s^2$

Substituting the value in formula -

$\implies$$\rm R = \frac{u^2sin2\theta}{g}$

$\implies$$\rm R = \frac{12^2 \times sin2(45)}{10}$

$\implies$$\rm R = \frac{144 \times 1 }{10}$

$\implies$$\rm R = \frac{144}{10}$

$\implies$$\rm R = 14.4 m$

The distance after which it will hit the ground again is 14.4 m

Answer 2

Explanation:

g=10.0 m/s²

u=12m/s

angle=45°

$range = \frac{{u}^{2} \sin(2 \alpha )}{g}$

R=144 X 1/10

R=14.4m