Projectile Motion (1 Viewer)

[Lukybear]

Find the speed and direction of a particle which , when projected form a point 15m above the horizontal ground, just clears the top of a wall 26.25 m and 30m away.

 

[Shikobe]

x'=Vcos0=V, x=Vt
y''=-g, y'=-gt, y=-1/2gt^2 + 15
sub, t=x/V into y, which gives: y=-1/2g(x^2/V^2)
sub x=30 and y=26.25 into the equation and you should get your answer.

Hope this helps.

 

[Lukybear]

theta's not zero.

But good try anywaz.

 

[k02033]

There is no unique solution to this problem,

Let T denote the time it takes for the projectile to reach horizontal displacement of 30m.

Let U denote initial speed, @ denote launch angle, a =-10.


Then:

Ucos@*T=30 (1)

Usin@T+1/2*a*T^2=6.25 (2)

put (1) into (2) and make @ the subject you get tan@=1/30*(6.25-1/2*a*T^2), so now we can solve @ in terms of T. And then use (1) to solve U in terms of T. So we can chose T=1,2.. or whatever appropreiate value as long as @ and U are still solvable, (so some values of T are not possible, e.g. T=0 makes U unsolvable)

in short there is no unique solution, since there are projectiles that can just clear that wall at diff times

 

[k02033]

Its pretty cool how mathematics translates into everyday logic, e.g. T=0 is not possible implies the projectile cant teleport. (Ignoring quantum mechanics of course >.>)

 

[Shikobe]

Oh sorry! I thought the particle was projected horizontally lol my bad.

i think the question is saying that at x=30 and y=26.25, it is the maximum height, so...

y=-1/2gt^2 + Vtsin@ + 15, y'=-gt+Vsin@, when y'=0, t=Vsin@/g,
Sub t into y: y=......(sorry getting lazy)
Then you put x=30 and y=26.25 in the equation above and you should get Vsin@=15

Then you know now the time for t at that point. t=Vsin@/g=15/10=3/2.

You can sub the value into x=Vtcos@, and you also know that x=30 at that time so...

30=3/2Vcos@, therefore Vcos@=20.

you divide y' and x' you get: tan@=15/20, @=36*52*

and V=sqrt(20^2+15^2)=25m/s

 

[hscishard]

Its pretty cool how mathematics translates into everyday logic, e.g. T=0 is not possible implies the projectile cant teleport. (Ignoring quantum mechanics of course >.>)

Lol like a particle moving x=1/t-1
Flys out of the universe and comes back down to the ground. (well hovers on the ground)

 

[mirakon]

Use equation of path

y= xtan theta + gx^2/(V^2cos^2 theta) + 15 (note: I added 15 as the particle is shot from 15m high)

where theta is angle of launch
V is launch speed
g is acceleration due to gravity.

sub the height of wall as y and distance away it is as x and then solve for theta