Question

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

Answer 1

Hi friend,

your answer

Answer 2

Explanation:

We know that,

$\rm Horizontal \: range = \dfrac{u^2\sin 2 θ}{g}$

$\rm = \dfrac{(12)^2×\sin (2×45^{\circ})}{10}$

$\rm = \dfrac{144}{10}$

$\rm = 14.4 \: m$

$\bf{Final \: answer }= \rm14.4 \: m$