Bobbert said:3. The Attempt at a Solution
V22=V12+2ad
0=(0.714sin30o)2+2(-9.81)(d)
d=19.62-0.127449
d=19.49m
Now this answer is not practical at all...
The max height of a projectile refers to the highest point reached by an object that is launched or thrown into the air and follows a parabolic path.
The max height of a projectile can be calculated using the equation: h = (v2 * sin2θ) / 2g, where h is the max height, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.
The max height of a projectile is affected by the initial velocity, launch angle, and acceleration due to gravity. Air resistance can also play a role in altering the max height.
Yes, there is an optimal launch angle for achieving the max height of a projectile. This angle is 45 degrees, as it allows for the most efficient conversion of initial velocity into vertical velocity.
Yes, the max height of a projectile can be greater than the initial height. This occurs when the launch angle is greater than 45 degrees, allowing for a longer time in the air and therefore a greater max height.