# Prove that a projectile launched at 45 degrees would travel the farthest.

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### 1 Answer

If a projectile is launched at an angle L to the horizontal, its initial velocity V can be divided into a component along the x-axis denoted by Vx and a component along the y-axis denoted by Vy.

Let the time the projectile is in motion be given by T. Now an acceleration g due to the gravitational force of attraction acts on the projectile during its motion. The acceleration is in the opposite direction to Vy. The projectile falls on the ground when the velocity imparted due to the gravitational attraction is equal to Vy but in the opposite direction. So, we have –Vy = Vy – g*T

=> 2*Vy = g*T

=> T = 2*Vy/g

As Vy = V sin L

=> T = (2*V*sin L)/g

Now in the time T if the projectile travels a horizontal distance D

D = Vx*T

=> D = Vx*(2*V*sin L)/g

Now Vx = V cos L

=> D = V* cos L*(2*V*sin L)/g

=> D = V^2 * 2* cos L* sin L/g

Using the relation 2 sin L* cos L = sin 2L

=>D = V^2 *sin 2L/g

Now sin 2L can take a maximum value of 1, and for this is to happen L = 45 degrees.

**Therefore a projectile launched at an angle of 45 degree travels the largest horizontal distance.**