Question

A projectile is launched at ground level with an initial speed of$\mathbf{50}\mathbf{.}\mathbf{0}\mathit{m}\mathbf{/}\mathit{s}$at an angle of$\mathbf{30}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}$above the horizontal. It strikes a target above the ground$\mathbf{3}\mathbf{.}\mathbf{00}$seconds later. What are the x and y distances from where the projectile was launched to where it lands?

Short answer

The projectile strikes a target at a distance of $\mathbf{130}$meters horizontally and $\mathbf{30}\mathbf{.}\mathbf{9}$meters vertically from the launching point.

Step-by-step solution

Step 1: Determine projectile Motion

The projectile is a missile that moves freely under the influence of gravity and air resistance after being propelled by an external source.

A projectile is an object that is traveling in both the horizontal and vertical directions at the same time. Projectile motion is always in the form of a parabola.

Given Data:

• Initial speed = 50.0 m/s
• Angle above the horizontal =${30}^{\circ }$
• Time = 3 s.

Step 2: Determine components of initial velocity

Since we do not know the exact location of the projectile after $3$s.

Components of the initial velocity can be calculated as:

$\begin{array}{rcl}\left(V_{ox}\right)& =& \left(50\right)\cos {30}^{°}\\ & =& 43.3013\mathrm{m}/\mathrm{s}\\ \left(V_{oy}\right)& =& \left(50\right)\sin {30}^{°}\\ & =& 25\mathrm{m}/\mathrm{s}\end{array}$

Step 3: To solve the values of x and y:

Horizontal displacement X can be calculated as:

$\mathit{X}\mathbf{=}{\mathit{v}}_{\mathbf{o}\mathbf{x}}\mathit{t}$

Substituting values in the above expression, we get,

$\begin{array}{rcl}X& =& (43.3013)\times \left(3\right)\\ X& =& 130\text{m}\\ & & \end{array}$

Vertical displacement Y can be calculated as:

$Y=v_{oy}t+\frac{1}{2}a{t}^2$

Substituting values in the above expression, we get,

$\begin{array}{rcl}Y& =& \left(25\text{m/s}\right)\times \left(3\text{s}\right)+\frac{1}{2}\times \left(-9.8\right)\times {\left(3\text{s}\right)}^2\\ Y& =& 30.9\text{m}\\ & & \end{array}$

Therefore, the projectile strikes a target at a distancerole="math" localid="1668679240901" $\mathbf{130}\mathbf{}\mathit{m}$ horizontally and $\mathbf{30}\mathbf{.}\mathbf{9}\mathbf{}\mathit{m}$vertically from the launching point.