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.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Projectile Motion: Positive Launch concept. If you need more Projectile Motion: Positive Launch practice, you can also practice Projectile Motion: Positive Launch practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Abushanab's class at CSUSB.