Aprojectile is launched from ground level with an initial speed of 41.5 m/s at an angle of 32.3° above the horizontal. it strikes a target in the air 2.22 seconds later. what are the horizontal and vertical distances from where the projectile was launched to where it hits the target?

To answer this question it must be taken into account that according to the kinematics equations for the position of an object that performs this type of movement is:

Where:

X (t) is the position of the object in the horizontal direction as a function of time

Y (t) is the position of the object in the vertical direction as a function of time

α is the angle of inclination in which the object is thrown

is the initial velocity of the object

g is the acceleration of gravity

t is the time

We want to know what distance the projectile traveled in 2.22 seconds after launch.

Then it is:

 After 2.22 seconds, the projectile strikes with the target at a horizontal distance of 77.87 m and a vertical distance of 25.08 m.

the median here is 25. your lowest value is 14 and highest value is 48. to find the third quartile, find the medium number in between 25 and 48 which would be 42 i believe.

x equals 1

step-by-step explanation:

f(4)=f(×)

x=1