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

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

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Answer 1 · 2017-05-29T06:44:41

The distance is $= 3.65 m$ $=3.65m$

Explanation:

Resolving in the vertical direction ${↑ ⏐ ⏐}^{+}$ $uarr^+$

initial velocity is $u_{y} = v sin θ = 8 \cdot sin (\frac{1}{4} π)$ $u_y=vsintheta=8*sin(1/4pi)$

Acceleration is $a = - g$ $a=-g$

At the maximum height, $v = 0$ $v=0$

We apply the equation of motion

$v = u + a t$ $v=u+at$

to calculate the time to reach the greatest height

$0 = 8 sin (\frac{1}{4} π) - g \cdot t$ $0=8sin(1/4pi)-g*t$

$t = \frac{8}{g} \cdot sin (\frac{1}{4} π)$ $t=8/g*sin(1/4pi)$

$= 0.577 s$ $=0.577s$

The greatest height is

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

Resolving in the horizontal direction $\to^{+}$ $rarr^+$

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

$s = u_{x} \cdot t$ $s=u_x*t$

$= 8 cos (\frac{1}{4} π) \cdot 0.577$ $=8cos(1/4pi)*0.577$

$= 3.27 m$ $=3.27m$

The distance from the starting point is

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

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

$= 3.65 m$ $=3.65m$

Answer 2 · 2017-05-29T08:59:48

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

Explanation:

$Projectile motion is a special case of two-dimensional motion.$ $"Projectile motion is a special case of two-dimensional motion. "$
$An object found at the moment of shooting is seen.$ $"An object found at the moment of shooting is seen."$

enter image source here

$if we ignore the air resistance and gravity,the object move to infinity.$ $"if we ignore the air resistance and gravity,the object move to infinity."$
$But that's not really the case. The object is attracted by the ground$ $"But that's not really the case. The object is attracted by the ground"$ $and placed on a trajectory.$ $" and placed on a trajectory."$
$The object drops freely from point J to point H.$ $"The object drops freely from point J to point H."$

enter image source here

$We must divide the motion horizontally and vertically into two parts.$ $"We must divide the motion horizontally and vertically into two parts."$

The blue vector shows the horizontal component of the initial velocity.

enter image source here

$The V_{x} vector can be calculated using v_{x} = v i \cdot cos θ$ $"The " V_x " vector can be calculated using " v_x=vi*cos theta$

The magnitude and direction of blue vector does not change.
*The blue vector allows the object to move horizontally. *

enter image source here

The green vector shows the vertical component of the initial velocity.

enter image source here

$The V_{y} vector can be calculated using v_{y} = v i \cdot sin θ$ $"The " V_y " vector can be calculated using " v_y=vi*sin theta$

$V_{y} at any time is v_{y} = v_{i} \cdot sin θ - g \cdot t$ $V_y " at any time is "v_y=v_i*sin theta -g*t$

enter image source here

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.

$How can I calculate the elapsed time from initial point to peak point ?$ $"How can I calculate the elapsed time from initial point to peak point ?"$

$v_{y} = 0 (at maximum height)$ $v_y=0("at maximum height)"$
$0 = v_{i} \cdot sin θ - g \cdot t$ $0=v_i*sin theta-g*t$
$v_{i} \cdot sin θ = g \cdot t$ $v_i*sin theta=g*t$
$t = \frac{v_{i} \cdot sin θ}{g}$ $t=(v_i*sin theta)/g$

$How can I calculate the maximum height?$ $"How can I calculate the maximum height?"$

enter image source here

$h_{m} = \frac{v_{i}^{2} \cdot {sin}^{2} θ}{2 \cdot g}$ $h_m=(v_i^2*sin^2 theta)/(2*g)$

$How can I calculate the elapsed time for traveled from initial point to end point ?$ $"How can I calculate the elapsed time for traveled from initial point to end point ?"$

$t = 2 \cdot \frac{v_{i} \cdot sin θ}{2}$ $t=2*(v_i*sin theta)/2$

$How do we find the velocity of a projectile at any time over its trajectory?$ $"How do we find the velocity of a projectile at any time over its trajectory?"$

enter image source here

$Calculate v_{x} = v_{i} \cdot cos θ$ $"Calculate "v_x=v_i*cos theta$
$Calculate v_{y} = v_{i} \cdot sin θ - g \cdot t$ $"Calculate "v_y=v_i*sin theta-g*t$
$Calculate v = \sqrt{{(v_{x})}_{x}^{2} + {(v_{y})}_{y}^{2}}$ $"Calculate "v=sqrt((v_x)^2+(v_y)^2)$

$How can I calculate the location of object on trajectory?$ $"How can I calculate the location of object on trajectory?"$

enter image source here

$x = v_{i} \cdot t \cdot cos θ$ $x=v_i*t*cos theta$
$y = v_{i} \cdot t \cdot sin θ - \frac{1}{2} g t^{2}$ $y=v_i*t*sin theta-1/2 g t^2$

$How can I calculate the maximum x-range?$ $"How can I calculate the maximum x-range?"$

$x_{m} = \frac{v_{i}^{2} \cdot sin (2 θ)}{g}$ $x_m=(v_i^2*sin(2 theta))/g$

$answer to your question.$ $"answer to your question."$
$θ = \frac{π}{4}$ $theta=pi/4$
$v_{i} = 8 m \cdot s^{- 1}$ $v_i=8 m*s^(-1)$

$t = \frac{v_{i} \cdot sin θ}{g} = \frac{8 \cdot 0.707}{9.81} = 0.58 s$ $t=(v_i*sin theta)/g=(8*0.707)/(9.81)=0.58 s$

$x = v_{i} \cdot t \cdot cos θ = 8 \cdot 0.58 \cdot 0.707 = 3.28 m$ $x=v_i*t*cos theta=8*0.58*0.707=3.28m$

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

2 Answers

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