For a given launch velocity, at which launch angle does the projectile undergo the maximum horizontal displacement?
A. 35°
B. 45°
C. 75°
D. 80°
E. 90°
L=vₒ²•sin2α/g,
B. 45°
To determine the launch angle at which the projectile undergoes the maximum horizontal displacement, we can analyze the projectile motion and use the concept of range.
The range of a projectile refers to the horizontal distance traveled by the projectile before hitting the ground. It is given by the equation:
Range = (v^2 * sin(2θ)) / g
where:
v is the launch velocity of the projectile,
θ is the launch angle, and
g is the acceleration due to gravity.
To find the launch angle that gives the maximum range, we can differentiate the range equation with respect to θ and set it equal to zero. This will help us find the maximum point.
Let's calculate the derivatives:
d(Range)/dθ = (v^2 * 2 * cos(2θ)) / g
Setting the derivative equal to zero:
(v^2 * 2 * cos(2θ)) / g = 0
From this equation, we can see that cos(2θ) = 0, which means that 2θ = 90° or θ = 45°.
Therefore, the launch angle at which the projectile undergoes the maximum horizontal displacement is 45°.
The correct option is B. 45°.