Question related to projectile motion?

Question

Can anybody calculate the the length traced by a projectile, shot with a speed of 5m/s and at an angle of 45 with the horizontal? one who solves this get the mathematics award :) ,but yes steps should b there!!

The length of the trajectory

Answer 1

You';re looking for an integral over the length of the arc. This is within 1st year college mathematics.

First, you need to derive an equation for the arc.

gravity is 9.8 m/sec^2.

First of all, we need to get with radians.

only sin(45) goes into height. So its upward velocity is 5*(sqrt(2)/2). That's about 3.5

In about 1/3rd of a second, it stops going up. In 2/3rds of a second it hits the ground.

Its horizontal displacement is 3.5 m/sec * 2/3rds of a second. This is, oh, 2.3 meters?

You can do these calculations more accurately, I'm working from memory in my head for all this.

So you have the points (0,0), (3.5* 1/3, 3.5*1/3) and (3.5*2/3, 0)

where x is the horizontal displacement, and y is the vertical displacement. Use your curve-fitting to plot a parobola through those points. Now, do your intergration over the line, again from your calculus textbook. Wasn't that easy?

I'm not sure how accurate you need to be. I think I'm within an order of magnitude, and there probably is an easier way to solve this. You should do the calculations with a calcuator.

Answer 2

Ignoring air resistance.
x(t) = 5*sin(45) * t ~ 3.5355 * t
y(t) = 5*cos(45) * t - 0.5 * 9.8 * t^2 ~ 3.5355*t - 4.9 * t^2

launch time at t = 0 s
lands when y(t) = 0
3.5355*t - 4.9 * t^2 = 0
t * ( 3.5355 - 4.9*t) = 0
4.9t = 3.5355
t = 0.7215 s

arclength (s) =
t
⌠ √ ( (dx/dt)^2 + (dy/dt)^2 ) dt
⌡
t0

dx/dt = 3.5355
(dx/dt)^2 = 12.5

dy/dt =3.5355 - 9.8t
(dy/dt)^2 = 12.5 - 34.65*t + 96*t^2

s =
0.7215
⌠ √(25 - 34.65*t + 96*t^2) dt
⌡
0

Getting late. I don't have the time to do this integration right now. I might get back to it later. or someone else can do it. I'm sure someone knows a good online integral solver.

Answer 3

Ignoring air resistance.
vy=vy0 + at
t=(vy-vy0)/a = -vy0/a = -3.535/-9.8 = 0.360768 sec
sy=vt+(at^2)/2 = 3.535*..360768 -4.9*t^2 = 0.637562 m

sx=3.535*0360768=1.27532 m

total distance traveled horizontally =2.55064 m

track is a parabola described by y-0.637562=-0.392(x-1.27532)^2, y=0.637562-0.392(x-1.27532)^2


y=y0 + vy0t + (at^2)/2, dy=vy0dt+atdt
x=vx0t, dx=vx0dt

ds^2 =dx^2 +dy^2 = vx0^2dt^2+(vy0+at)^2dt^2



let y=ax^2
dy = 2xdx
ds^2 = 4a^2x^2dx^2+dx^2=(4a^2x^2+1)dx^2
ds = dx/(1+4a^2x^2)^.5
s/2 = (arcsinh(2ax))/2a for half the parabola,
y=0.637562-0.392(x-1.27532)^2
s = 4.496 m