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※ 引述《rabbit14175 (豬豬火)》之銘言：
: question 1
: A rifle pointed at 30? to the horizontal fires a bullet at 250m/s.
: If the bullet is accelerated uniformly in the barrel for 0.006s,
: find
: (a) the magnitude of the acceleration;
: (b) its horizontal and vertical components.
: question 2
: Show that the horizontal and vertical displacement components for
: a projectile satisfy an equation of the form △y=a△x+b(△x)^2, so
: that the trajectory is a parabola.
: 小弟不才...煩請物理高手解答
Well, you know
As a projectile is rising, we have
{ △y = Vy*t - 1/2*gt^2 ----eq.1
{ △x = Vx*t ----eq.2
eq.2 => t = △x/Vx ----eq.3
replace the "t" of eq.1 with eq.3
Thus, △y = (Vy/Vx)△x - 1/2*g(△x/Vx)^2
= -1/2*(g/Vx)^2*(△x)^2 + (Vy/Vx)△x
--> This minus means
that the parabola is downward as rising.
Similarly, as a projectile is dropping off, we have
{ △y = 1/2*g(T)^2 - 1/2*gt^2 ----eq.1
{ △x = Vx*(T+t) ----eq.2
{ where T is the rising time
{ T = Vy/g ----eq.3
replace T of eq.1 and eq.2 with eq.3
=> { △y = 1/2*g(Vy/g)^2 - 1/2*gt^2 ----eq.1'
{ △x = Vx*(Vy/g+t) ----eq.2'
eq.2' => t = (△x/Vx) - Vy/g ----eq.2"
replace the "t" of eq.1' with eq.2"
Thus, △y = 1/2*g(Vy/g)^2 - 1/2*g[(△x/Vx) - Vy/g]^2
= 1/2*g(Vy/g)^2 - 1/2*g[(△x/Vx)^2 -2(△x/Vx)Vy/g
+ (Vy/g)]^2
= -1/2*(g/Vx)^2*(△x)^2 + (Vy/Vx)△x
--> This minus means
that the parabola is downward as dropping.
Finally, △y = -1/2*(g/Vx)^2*(△x)^2 + (Vy/Vx)△x
it's a parabola.
---------------------------------------------
I am not an expert.
I just understand it early than you.
※ 編輯: moleming 來自: 140.112.101.46 (10/20 22:27)
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