# A projectile is shot from the ground at an angle of pi/4  and a speed of 8 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

May 29, 2017

The distance is $= 3.65 m$

#### Explanation:

Resolving in the vertical direction ${\uparrow}^{+}$

initial velocity is ${u}_{y} = v \sin \theta = 8 \cdot \sin \left(\frac{1}{4} \pi\right)$

Acceleration is $a = - g$

At the maximum height, $v = 0$

We apply the equation of motion

$v = u + a t$

to calculate the time to reach the greatest height

$0 = 8 \sin \left(\frac{1}{4} \pi\right) - g \cdot t$

$t = \frac{8}{g} \cdot \sin \left(\frac{1}{4} \pi\right)$

$= 0.577 s$

The greatest height is

$h = {\left(8 \sin \left(\frac{1}{4} \pi\right)\right)}^{2} / \left(2 g\right) = 1.63 m$

Resolving in the horizontal direction ${\rightarrow}^{+}$

To find the horizontal distance, we apply the equation of motion

$s = {u}_{x} \cdot t$

$= 8 \cos \left(\frac{1}{4} \pi\right) \cdot 0.577$

$= 3.27 m$

The distance from the starting point is

$d = \sqrt{{h}^{2} + {s}^{2}}$

$= \sqrt{{1.63}^{2} + {3.27}^{2}}$

$= 3.65 m$

May 29, 2017

Please check the animations carefully. The answer is given at the end of the explanation.

#### Explanation:

$\text{Projectile motion is a special case of two-dimensional motion. }$
$\text{An object found at the moment of shooting is seen.}$ $\text{if we ignore the air resistance and gravity,the object move to infinity.}$
$\text{But that's not really the case. The object is attracted by the ground}$$\text{ and placed on a trajectory.}$
$\text{The object drops freely from point J to point H.}$ $\text{We must divide the motion horizontally and vertically into two parts.}$

• The blue vector shows the horizontal component of the initial velocity. $\text{The " V_x " vector can be calculated using } {v}_{x} = v i \cdot \cos \theta$

• The magnitude and direction of blue vector does not change.
• *The blue vector allows the object to move horizontally. * • The green vector shows the vertical component of the initial velocity. $\text{The " V_y " vector can be calculated using } {v}_{y} = v i \cdot \sin \theta$

${V}_{y} \text{ at any time is } {v}_{y} = {v}_{i} \cdot \sin \theta - g \cdot t$ • The magnitude and direction of green vector change.
• *The blue vector allows the object to move vertically. *
• the magnitude of green vector at maximum height is zero.

$\text{How can I calculate the elapsed time from initial point to peak point ?}$

v_y=0("at maximum height)"
$0 = {v}_{i} \cdot \sin \theta - g \cdot t$
${v}_{i} \cdot \sin \theta = g \cdot t$
$t = \frac{{v}_{i} \cdot \sin \theta}{g}$

$\text{How can I calculate the maximum height?}$ ${h}_{m} = \frac{{v}_{i}^{2} \cdot {\sin}^{2} \theta}{2 \cdot g}$

$\text{How can I calculate the elapsed time for traveled from initial point to end point ?}$

$t = 2 \cdot \frac{{v}_{i} \cdot \sin \theta}{2}$

$\text{How do we find the velocity of a projectile at any time over its trajectory?}$ $\text{Calculate } {v}_{x} = {v}_{i} \cdot \cos \theta$
$\text{Calculate } {v}_{y} = {v}_{i} \cdot \sin \theta - g \cdot t$
$\text{Calculate } v = \sqrt{{\left({v}_{x}\right)}^{2} + {\left({v}_{y}\right)}^{2}}$

$\text{How can I calculate the location of object on trajectory?}$ $x = {v}_{i} \cdot t \cdot \cos \theta$
$y = {v}_{i} \cdot t \cdot \sin \theta - \frac{1}{2} g {t}^{2}$

$\text{How can I calculate the maximum x-range?}$

${x}_{m} = \frac{{v}_{i}^{2} \cdot \sin \left(2 \theta\right)}{g}$

$\text{answer to your question.}$
$\theta = \frac{\pi}{4}$
${v}_{i} = 8 m \cdot {s}^{- 1}$

$t = \frac{{v}_{i} \cdot \sin \theta}{g} = \frac{8 \cdot 0.707}{9.81} = 0.58 s$

$x = {v}_{i} \cdot t \cdot \cos \theta = 8 \cdot 0.58 \cdot 0.707 = 3.28 m$