The formula HM=U2Sin2 2g...(1), where U is the initial velocity, and g is the acceleration due to gravity, determines the highest height that a projectile can reach while disregarding air resistance. (28.2m/s).
a) The initial velocity components of a projectile thrown at a speed of vi and an angle of I above the horizontal are vxi = vi cos I and vyi = vi.sin i. When the projectile returns to the level from which it was launched (in this case, the ground), ignoring air resistance, its vertical velocity will be v.y=-vyi.
With this knowledge, the entire flight time can be calculated as follows: t total = v.yf v yi/ay = v yi v yi/g = 2vyi/g or t total = 2vi sini/g.
Given that a projectile's horizontal velocity is constant in the absence of air resistance, its range may be calculated using the following formula:
R=vxi t total =(vi cosθ i)( 2vi sinθ i/g )= vi2/g (2sinθ i cosθ i )
= vi2 sin(2θ 1)/g.
Thus, if the projectile is to have a range of R=81.1m when launched at an angle of θ i =45.0 0 , the required initial speed is
vi= [tex]\sqrt{x} Rg/sin(2θ i )[/tex]= [tex]\sqrt{x} (81.1m)(9.80m/s 2 )/sin(90.0 0 )[/tex]=28.2m/s
The apex—the highest point on any trajectory—is attained when vy = 0. We can compute y because we know the initial position, initial velocity, and final velocity by using the equation v2y=v20y2g(yy0).
the angle at which a projectile's initial velocity and the horizontal axis coincide. These angles are often 90 degrees or smaller.
Learn more about projectile here:
https://brainly.com/question/11422992
#SPJ4