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Q25PE

Expert-verifiedFound in: Page 123

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A projectile is launch**ed at gro**und 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?**

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.

**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.

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

Components of the initial velocity can be calculated as:

$\begin{array}{rcl}\left({V}_{ox}\right)& =& \left(50\right)\mathrm{cos}{30}^{\xb0}\\ & =& 43.3013\mathrm{m}/\mathrm{s}\\ \left({V}_{oy}\right)& =& \left(50\right)\mathrm{sin}{30}^{\xb0}\\ & =& 25\mathrm{m}/\mathrm{s}\end{array}$

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.

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