A projectile is shot from the ground at an angle of $\frac{π}{8}$ $pi/8$ and a speed of $\frac{2}{7} \frac{m}{s}$ $2 /7 m/s$ . 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 $\frac{π}{8}$ $pi/8$ and a speed of $\frac{2}{7} \frac{m}{s}$ $2 /7 m/s$ . 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

1683 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-15T13:52:24

There are several approaches to this type of question. The most direct method is to find time $t$ $t$ at the peak, when $v_{y} = 0$ $v_y =0$ , and use that time and $v_{x}$ $v_x$ to find the horizontal distance $d_{x}$ $d_x$ .

Explanation:

Projectiles follow a parabola, assuming (as most questions do) that air resistance is negligible. In a parabola, the slope is 0 at the vertex, which is the peak, or turnaround point between going up and coming down.

So any time a projectile question says "maximum height" you know that $v_{y} = 0$ $v_y = 0$ at that point.

To solve a projectile problem like this we need both horizontal and vertical components of the velocity, which we can get using trigonometry:

$v_{i y} = v_{i} sin (θ)$ $v_(iy) = v_i sin(theta)$
$v_{x} = v_{i} cos (θ)$ $v_x = v_i cos(theta)$
where $v_{i}$ $v_i$ is the initial velocity ( $\frac{2}{7} \frac{m}{s}$ $2/7m/s$ ) and $θ$ $theta$ is the angle ( $\frac{π}{8}$ $pi/8$ )
so
$v_{i y} = 0.109 \frac{m}{s}$ $v_(iy) = 0.109 m/s$
$v_{x} = 0.264 \frac{m}{s}$ $v_x = 0.264 m/s$

Using kinematics equations, we can find $t$ $t$

$a = \frac{v_{f y} - v_{i y}}{t}$ $a = (v_(fy)-v_(iy))/t$
$t = \frac{v_{f y} - v_{i y}}{a}$ $t = (v_(fy)-v_(iy))/a$ and $a$ $a$ is gravity, $- 9.8 \frac{m}{s^{2}}$ $-9.8 m/s^2$
$t = \frac{0 - 0.109}{- 9.8} = 0.0111 s$ $t = (0-0.109)/(-9.8) = 0.0111s$

now

$d_{x} = v_{x} t$ $d_x=v_xt$ , so
$d_{x} = 0.264 \cdot 0.0111 = .00294$ $d_x=0.264*0.0111 = .00294$

That's a pretty feeble projectile!

There is another solution, which is faster, but is more indirect. We can exploit the fact that parabolas are symmetrical, so we can also find the total horizontal range, and divide by 2.

There is a formula for horizontal range, but it must be used cautiously - it only applies when the start and end height are the same, ie a flat surface.

$d_{x} = \frac{v_{i}^{2} \cdot sin (2 θ)}{g}$ $d_x=(v_i^2*sin(2theta))/g$

$d_{x} = \frac{{(\frac{2}{7})}^{2} \cdot sin (2 (\frac{π}{8}))}{9.8} = .000589$ $d_x=((2/7)^2*sin(2(pi/8)))/9.8 = .000589$
$\frac{d_{x}}{2} = .00294$ $d_x/2 = .00294$

Science

Math

Humanities

... and beyond

A projectile is shot from the ground at an angle of $\frac{π}{8}$ $pi/8$ and a speed of $\frac{2}{7} \frac{m}{s}$ $2 /7 m/s$ . 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 π8pi/8 and a speed of 27ms2 /7 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 $\frac{π}{8}$ $pi/8$ and a speed of $\frac{2}{7} \frac{m}{s}$ $2 /7 m/s$ . Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?