Sunday, December 23, 2012

Finding the Velocity, Time and Distance for a Projectile Launched Above a Moving Target - Kind of Solved

a 1 cm projectile is launched at a 45° angle 100 m above a 1 cm target. the target is at a horizontal distance of 5 m from the vertical. just as the projectile is launched, the 1 cm target moves at a steady 9 m/s. there is no air resistance.

find the:

(1) velocity at which the projectile must be launched to hit the target
(2) maximum height of the projectile
(3) time it will take to hit the target
(4) horizontal distance where the hit will take place
(5) velocity of the projectile at the time of impact

Hx = initial horizontal distance of moving target
Hvx = velocity of moving target
v = velocity of projectile
v-impact = velocity at the time of impact
h = height projectile is launched from
α = angle
g = gravity at 9.8 m/s
t = time

we use the two equations which appear directly below the three sentences i'm about to type to find the velocity of the launched projectile. each equation represents the x-position of where the target and the projectile will meet. so far we have had only two sentences, but if i add another one we will have the three that i spoke of earlier in the first sentence of this paragraph. some would put a period after "two sentences" and then begin another sentence with but. however, that would make four sentences which is not the amount i promised there would be at the beginning. and anyway, it's too late now to go back and make the necessary changes (how many sentences have we had so far? i've lost count).

Hx + Hvx(t) = cosα v(t)

the following is the equation for time:

make an informed estimate, then do a trial and error.
sorry, but that’s the best i could do. hey, at least it works!

(if anyone can tell me how to algebraically incorporate the time equation into the above two equations to solve for velocity the way the good lord meant for it to be solved, i would appreciate it on a scale not seen in a fortnight)

plugging in the numbers we find:

v = 13.982 m/s
maximum height = 104.987 m
v-impact = 46.427 m/s 
t = 5.638 s
x = 55.739 m ≅ 55.7383 m

a difference of 0.7 mm. that’s pretty damn close for my liking.

maximum height = the top half of the first part of the equation for time.
v-impact = the square root of the square of Vx + Vy at the time of impact. Vx is cosα v and Vy can be found by multiplying the first part of the equation for time by 9.8.


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