A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0º above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

Solution:

Since we do not know the exact location of the projectile after 3 seconds, consider the following arbitrary figure:

From the figure, we can solve for the components of the initial velocity.

\begin{align*}
\text{v}_{\text{ox}} &=\left(50\:\text{m/s}\right)\cos 30^{\circ} \\
& =43.3013\:\text{m/s}
\\
\\
\text{v}_{\text{oy}} & =\left(50\:\text{m/s}\right)\sin 30^{\circ} \\
&=25\:\text{m/s}
\\
\end{align*}

So, we are asked to solve for the values of x and y. To solve for the value of the horizontal displacement, x, we shall use the formula x=v_oxt. That is,

\begin{align*}
\text{x} & =\text{v}_{\text{ox}}\text{t} \\
\text{x} & =\left(43.3013\:\text{m/s}\right)\left(3\:\text{s}\right) \\
\text{x} & =130\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

To solve for the vertical displacement, y, we shall use the formula y=v_oyt+1/2at². That is

\begin{align*}
\text{y} & =\text{v}_{\text{oy}}\text{t}+\frac{1}{2}\text{a}\text{t}^2 \\
\text{y} & =\left(25\:\text{m/s}\right)\left(3\:\text{s}\right)+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\left(3\:\text{s}\right)^2 \\
\text{y} & =30.9\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Therefore, the projectile strikes a target at a distance 129.9 meters horizontally and 30.9 meters vertically from the launching point.