Time taken for both the horizontal motion (which is the same as the time for the vertical motion) is expressed as:

$\overline{){\mathbf{t}}{\mathbf{=}}\frac{\mathbf{\Delta x}}{{\mathbf{v}}_{\mathbf{x}}}}$

The height along the vertical direction from which the rock was launched is given as:

$\overline{){{\mathit{y}}}_{\mathbf{f}}{\mathbf{-}}{{\mathit{y}}}_{{\mathbf{i}}}{\mathbf{=}}{{\mathbf{v}}}_{{\mathbf{0}}}{\mathbf{t}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{a}}{{\mathbf{t}}}^{{\mathbf{2}}}}$

**a)**

Get time from the first equation, as

t = 18.0/(10.5)cos(30.0°) = 1.979 s

a = - g = -9.810 m/s^{2}

A rock thrown with speed 10.5 m/s and launch angle 30.0° (above the horizontal) travels a horizontal distance of = 18.0 m before hitting the ground. Use the value = 9.810 for the free-fall acceleration.

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