※ 引述《rabbit14175 (猪猪火)》之铭言： : 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...烦请物理高手解答 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. ※ 编辑: moleming 来自: 140.112.101.46 (10/20 22:27)