That's almost the formula for the maximum height of a trajectory. To correct it, replace sin2θ with sinθ.staka said:I would like to simply know what Δy=[(sin2θ)(Vi)]^2/2a represents..
The formula only requires the angle and the initial velocity (along with the usual a=9.8m/s^2), so what height does the equation represent?
The equation for projectile motion is: y = y0 + v0t + 1/2at2, where y is the vertical position, y0 is the initial vertical position, v0 is the initial velocity, t is the time, and a is the acceleration.
Horizontal projectile motion involves motion in the x-direction, where the acceleration is zero and the velocity remains constant. Vertical projectile motion involves motion in the y-direction, where the acceleration is due to gravity and the velocity changes over time.
The range of a projectile can be found using the formula: R = v0cosθ * (v0sinθ + √(v0sinθ)2 + 2gy0) / g, where v0 is the initial velocity, θ is the launch angle, g is the acceleration due to gravity, and y0 is the initial vertical position.
Yes, the projectile motion formula can be used for non-uniform acceleration by breaking the motion into small intervals and using the formula for each interval. This can provide a more accurate result for complex scenarios.
Air resistance can affect projectile motion by slowing down the projectile's velocity and reducing its range. This is because air resistance creates a force in the opposite direction of motion, which decreases the projectile's acceleration and alters its trajectory.