College Physics by Openstax Chapter 3 Problem 25

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:

The path of the projectile from the ground to a point 3 seconds later.

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

vox=(50 m/s)cos⁡30∘=43.3013 m/svoy=(50 m/s)sin⁡30∘=25 m/s\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=voxt. That is,

x=voxtx=(43.3013 m/s)(3 s)x=130 m  (Answer)\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=voyt+1/2at2. That is

y=voyt+12at2y=(25 m/s)(3 s)+12(−9.81 m/s2)(3 s)2y=30.9 m  (Answer)\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.

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