The range R of a projectile is defined as the magnitude of the horizontal displacement of the projectile when it returns to its original altitude. (In other words, the range is the distance between the launch point and the impact point on flat ground.) A projectile is launched at t = 0 with initial speed vi at an angle θ above the horizontal. The range is

R =
vi2sin2θ
g

Use the expression to find the followings.
1. The maximum range of a projectile with launch speed vi.
a)vi^2/2g

b)vi^2/4g

c)4vi^2/g

d)2vi^2/g

e)vi^2/g

(b) The launch angle θ at which the maximum range occurs.

sin of anything is maximum when anything is 90 degrees or pi/2 radians. That maximum is 1.

Then
R = Vi^2/g times one
and 2 theta = 90 so theta = 45 degrees

U2 Sin 2a

= g
=34.66m

To find the maximum range of a projectile, we can start by using the expression for the range:

R = (vi^2 * sin(2θ)) / g

1. Maximum range of a projectile with launch speed vi:
To find the maximum range, we need to determine the launch angle (θ) that provides the maximum value for sin(2θ). We know that the maximum value for sin(2θ) is 1, which occurs when 2θ = 90° or θ = 45°.

Substituting θ = 45° into the range equation, we get:

R = (vi^2 * sin(90°)) / g
R = (vi^2 * 1) / g
R = (vi^2) / g

Therefore, the maximum range of a projectile with launch speed vi is:

R = vi^2 / g

So, the correct answer is (a) vi^2 / g.

(b) Launch angle θ at which the maximum range occurs:
We already found that the maximum range occurs when 2θ = 90° or θ = 45°. Therefore, the launch angle (θ) at which the maximum range occurs is 45°.

Therefore, the correct answer is 45°.