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Found in: Page 123

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

A projectile is launched at ground level with an initial speed of $\mathbf{50}\mathbf{.}\mathbf{0}\mathbit{m}\mathbf{/}\mathbit{s}$ 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?

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.

See the 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:

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

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

Horizontal displacement X can be calculated as:

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

Substituting values in the above expression, we get,

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

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,

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

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