Physics Question?

Question

A projectile can be fired at many angles between zero degrees (horizontally) and ninety degrees (vertically). Which angle would give the greatest horizontal displacement down range? Which pairs of angles give the same landing spot?

Please show/explain all work. Please show what formulas were used.

Answer 1

45º == greatest horizontal displacement

30º and 60º == same landing spot

Answer 2

range=Vo^2sin(2theta)/g


The largest value of the sine function is 1. For this function, it's 1 when theta is 45 degrees.

Answer 3

The equations of motion are
vy(t)=vi*sin(θ)-g*t
y(t)=vi*sin(θ)*t-.5*g*t^2
since the time to apogee is 1/2 the total flight time
0=vi*sin(θ)-g*t
t=vi*sin(θ)/g
for range
2*t is impact

x(t)=vi*cos(θ)*t
for impact
x(impact)=2*sin(θ)*cos(θ)*vi^2/g

using the identity that
2*sin(θ)*cos(θ)=sin(2*θ)

x(impact)=sin(2*θ)*vi^2/g
taking vi^2/g to be the constant K

x(impact)=sin(2*θ)*K

Now, to find the angle for the max range, take
dx/dθ and set equal to 0
using the chain rule

0=2*k*cos(2*θ)
cos(x)=0 when x=90

2*θ=90
θ=45

Now the pairs of angles that give the same range

again
x(impact)=sin(2*θ)*K

note that
sin(2*θ) is a periodic function
for
0<=θ<=90
sin(2*θ)=sin(2*(90-θ))
so for positive x range the pairs are in the range
0<=θ<=90
and
θ, (90-θ)

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