How Does Projectile Angle Affect Range in Physics Calculations?

[CrankFan]

In section 13.4 of Stewart's Calculus: Early Transcendentals, an example problem is worked out. I always seem to have problems grasping the physics stuff. In the example below I'll point out the spot where I can no longer follow.

(LaTeX codes don't seem to be working for me at the moment so I'll do this in ascii and revise with LaTeX when/if possible.)

A projectile is fired with an angle of elevation alpha and initial velocity v_0. Assuming that the air resistance is negligible and the only external force is due to gravity, find the possition function r(t) of the projectile. What value of alpha maximizes the range?

Solution: We set up the axes so that the projectile starts at the origin. Since the force due to gravity acts downward , we have

F = ma = -mgj

Where g = |a| ~= 9.8 m/s^2. Thus

a = -gj

Since v'(t) = a, we have

v(t) = -gtj + C

Where C = v(0) = v_0. Therefore

r'(t) = v(t) = -gtj + v_0

Integrating again, we obtain

r(t) = -1/2 gt^2 j + tv_0 + D

But D = r(0) = 0, so the position vector of the projectile is given by

[3] r(t) = -1/2 gt^2j + tv_0


***** I follow everything up to here fine, but the next step is where I get lost. *****

If we write |v_0| = v_0 (the initial speed of the projectile), then

v_0 = v_0 cos(alpha)i + v_0 sin(alpha)j

 

[amcavoy]

In section 13.4 of Stewart's Calculus: Early Transcendentals, an example problem is worked out. I always seem to have problems grasping the physics stuff. In the example below I'll point out the spot where I can no longer follow.

(LaTeX codes don't seem to be working for me at the moment so I'll do this in ascii and revise with LaTeX when/if possible.)

A projectile is fired with an angle of elevation alpha and initial velocity v_0. Assuming that the air resistance is negligible and the only external force is due to gravity, find the possition function r(t) of the projectile. What value of alpha maximizes the range?

Solution: We set up the axes so that the projectile starts at the origin. Since the force due to gravity acts downward , we have

F = ma = -mgj

Where g = |a| ~= 9.8 m/s^2. Thus

a = -gj

Since v'(t) = a, we have

v(t) = -gtj + C

Where C = v(0) = v_0. Therefore

r'(t) = v(t) = -gtj + v_0

Integrating again, we obtain

r(t) = -1/2 gt^2 j + tv_0 + D

But D = r(0) = 0, so the position vector of the projectile is given by

[3] r(t) = -1/2 gt^2j + tv_0


***** I follow everything up to here fine, but the next step is where I get lost. *****

If we write |v_0| = v_0 (the initial speed of the projectile), then

v_0 = v_0 cos(alpha)i + v_0 sin(alpha)j

It is just the process of taking a speed and representing it as a velocity, provided there is enough information. The magnitude of the velocity (speed) times the cosine of the angle will give you the speed in the x direction while multiplying it by the sine will give you the speed in the y direction. Writing these in vector form, you now have a velocity vector.

Alex

 

[CrankFan]

Ok, I think I get it. v_0 / |v_0| is the unit tangent vector and when the object has traveled one unit we know that x = cos(alpha) and y = sin(alpha), so

v_0 / |v_0| = cos(alpha)i + sin(alpha)j

and

v_0 = |v_0| cos(alpha)i + |v_0| sin(alpha)j

and in general we can represent the initial velocity vector in terms of initial angle and initial speed, as above.

Thanks, I'm slow to absorb this stuff.

 

[HallsofIvy]

Or just think of it as trigonometry: represent the initial velocity vector as a line of length v_0 at angle alpha to the x-axis. Drawing in the x and y components, you have a right triangle with hypotenuse v_0 and angle alpha. sin(alpha)= "opposite side over hypotenuse"= v_y/v_0 so v_y= v_0 sin(alpha)
cos(alpha)= "near side over hypotenuse"= v_x/v_0 so v_x= v_0 cos(alpha).