why does a projectile of 45 degrees have the greatest range?
recall that the formula for range is:
R = (vo)^2 * sin (2*theta) / (2g)
where
vo = initial velocity
theta = angle of release
g = acceleration due to gravity (9.8 m/s^2)
now, if we're given the initial velocity, the value of range will depend on theta,, observing this, if we want to have a max Range, we have to find the max value of sin (2*theta),, but the max value of sine is always 1, therefore, equating this to 1:
sin (2*theta) = 1
2*theta = arcsin 1
2*theta = 90 degrees
theta = 45 degrees
hope this helps~ :)
oops, sorry the formula for range must be
R = (vo)^2 * sin (2*theta) / g
i just remembered that the 2g is for the max height:
h,max = (vo)^2 * [sin (theta)]^2 / 2g
but explanation and proving is still the same~ :)
the greatest sin value of theta is the sin of 45*
A projectile at a 45-degree angle has the greatest range because of a combination of two factors: the horizontal and vertical components of its velocity.
To understand this, let's break down the motion of the projectile. When a projectile is launched, it follows a curved path in the shape of a parabola. This path is a result of the interaction between the initial velocity and the force of gravity.
The initial velocity of the projectile can be separated into two components: the horizontal component and the vertical component. The horizontal component determines how far the projectile travels horizontally, while the vertical component influences the height the projectile reaches and the time it takes to hit the ground.
At a 45-degree angle, the initial velocity is equally divided into its horizontal and vertical components. This means that the projectile has the maximum horizontal velocity and an ideal combination of vertical velocity and time of flight.
When the projectile is launched at angles other than 45 degrees, the range decreases. For angles less than 45 degrees, the vertical component of the velocity becomes more significant, allowing the projectile to reach higher altitudes but sacrificing horizontal distance. On the other hand, for angles greater than 45 degrees, the horizontal component of the velocity decreases, resulting in a shorter range.
To calculate the maximum range of a projectile launched at 45 degrees, you can consider the initial velocity, the acceleration due to gravity, and other factors such as air resistance, if applicable. By using the equations of motion and kinematics, you can determine the range mathematically.