A projectile is shot from the ground at an angle of $\frac{π}{4}$ $pi/4$ and a speed of $5 \frac{m}{s}$ $5 m/s$ . Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Question

A projectile is shot from the ground at an angle of $\frac{π}{4}$ $pi/4$ and a speed of $5 \frac{m}{s}$ $5 m/s$ . Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

1 Answer

Impact of this question

2000 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-27T22:21:00

Find the time it takes to reach the maximum height, and use the x component of the displacement equation to find the distance travelled in that time.

Explanation:

First we break up the velocity into it's vector components at time 0, $v_{x} (0)$ $v_x(0)$ and $v_{y} (0)$ $v_y(0)$ by using the identities:

$v_{x} (0) = v (0) cos (θ)$ $v_x(0)=v(0)cos(theta)$
$v_{y} (0) = v (0) sin (θ)$ $v_y(0)=v(0)sin(theta)$

Since we're presumably ignoring air resistance, the acceleration components are

$a_{x} = 0$ $a_x=0$
$a_{y} = - g$ $a_y= -g$

So our functions for $v_{x} (t)$ $v_x(t)$ and $v_{y} (t)$ $v_y(t)$ becomes

$v_{x} (t) = v (0) cos (θ)$ $v_x(t)=v(0)cos(theta)$
$v_{y} (t) = v (0) sin (θ) - g \cdot t$ $v_y(t)=v(0)sin(theta)-g*t$

We can find the time it takes to reach the maximum height by setting $v_{y} (t) = 0$ $v_y(t)=0$ which gives:

$t = v (0) \frac{sin (θ)}{g}$ $t=v(0)sin(theta)/g$

Then we use the displacement formula

$x (t) = x (0) + v_{x} t + \frac{1}{2} a_{x} t^{2}$ $x(t)=x(0)+v_x t+1/2 a_xt^2$

Since $a_{x} = 0$ $a_x=0$ , and $x (0) = 0$ $x(0)=0$ this simplifies to

$x (t) = v_{x} t$ $x(t)=v_x t$

Substitution of $v_{x}$ $v_x$ yields

$x (t) = v (0) cos (θ) t$ $x(t)=v(0)cos(theta) t$

and substituting for t yields:

$x (t) = \frac{v (0) cos (θ) v (0) sin (θ)}{g}$ $x(t)=(v(0)cos(theta)v(0)sin(theta))/g$ which simplifies to

$x (t) = \frac{v {(0)}^{2} cos (θ) sin (θ)}{g}$ $x(t)=(v(0)^2cos(theta)sin(theta))/g$ and further simplifies to

$x (t) = \frac{v {(0)}^{2} sin (2 θ)}{2 g}$ $x(t)=(v(0)^2sin(2theta))/(2g)$

This gives the range of the projectile at the maximum height.

It's interesting to note that this is exactly half the range equation for a projectile returning to the same height.

$R = \frac{v {(0)}^{2} sin (2 θ)}{g}$ $R=(v(0)^2sin(2theta))/(g)$

which is derived using

$y (t) = y (0) + v_{y} t - \frac{1}{2} g t^{2}$ $y(t)=y(0)+v_y t-1/2g t^2$

which is the displacement equation for y.

Science

Math

Humanities

... and beyond

A projectile is shot from the ground at an angle of $\frac{π}{4}$ $pi/4$ and a speed of $5 \frac{m}{s}$ $5 m/s$ . Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A projectile is shot from the ground at an angle of pi/4 π4pi/4 and a speed of 5 m/s5ms5 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?

1 Answer

Explanation:

Related questions

Impact of this question

A projectile is shot from the ground at an angle of $\frac{π}{4}$ $pi/4$ and a speed of $5 \frac{m}{s}$ $5 m/s$ . Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?