Question #ee158

1 Answer
Jan 24, 2018

Starting at 0, increasing the angle up to pi/4 (45^@) will increase the distance the object travels. After this point up to pi/2 (90^@), increasing the angle decreases the distance travelled.

Explanation:

This does intuitively make sense. 45^@ is the angle half-way from 0 and a right angle, so this should maximise both vertical distance and horizontal height.

Nonetheless, we can also prove this using a bit of algebra and calculus. Even if you can't do the second bit, first bit should make sense. The task is:

Consider this diagram.
Online homework questionOnline homework question
Prove that the distance r=-(2u^2sinthetacostheta)/g









Still here? Lets go.

Let's say we are firing a projectile with a velocity of u at an angle theta, and we want to find the horizontal range r. The particle is accelerating to the ground with accelerating g.
Considering vectors, the vertical velocity of the particle is usintheta.

Find the time it takes for the particle to return to the ground if we throw it up and that velocity.
s=0 It's returning to the start.
u=usintheta
v="/" we don't care what v is
a=-g since we've chosen up to be the positive direction, this means down must be negative.
t=t this is what we want to find.

Using the equation s=ut+1/2at^2

0=utsintheta+1/2g t^2
0=1/2t(g t+2usintheta)
t=(-2usintheta)/g

Now, consider the horizontal motion

s=r this is what we want to find
u=ucostheta
v="/"
a=0
t=-(2usintheta)/g

since a=0, s=ut

r=-ucostheta*(2usintheta)/g
r=-(2u^2sinthetacostheta)/g

Now, as an extra challenge, can you prove that the angle that gives the maximum r is pi/4 (45^@)?




We can use some trig identities to help us with this expression, then differentiate to find the maximum angle.

Since 2sinxcosx=sin2x;

r=-u^2/gsin(2theta)

Now, differentiate both sides wrt theta. This gives us the gradient function. When this is equal to zero, we will find the maximum value.
Since d/dx(sinkx)=kcoskx;

(dr)/(d theta)=-(2u^2)/gcos(2theta)

Let (dr)/(d theta)=0

-(2u^2)/gcos(2theta)=0

cos(2theta)=0

We want an angle between 0 and a right angle, so we want a value in the range 0<=theta<=pi/2

2theta=pi/2" or " 90^@

theta=pi/4" or " 45^@