A projectile is fired with an initial speed of 30 m/s at an angle of elevation of 30°. Find the following: anionalco at which the maximum height is achieved is The maximum height achieved by the projectile is The time when the projectile hits the ground is 앓 . The range of the projectile is m. The speed of the projectile on impact with the ground is 嚣| m/s.

1) Time to reach maximum height, t = 1.531s

2) Maximum height reached, H = 11.48m

3) Total time of flight, T = 3.062s

4) Range of the projectile, R = 79.53m

5) Speed with which the projectile hits the ground = 30 m/s

Step-by-step explanation:

Given,

U = 30m/s, θ = 30°, g = 9.8 m/s²

For projectile motion, the formulas for the parameters of this motion are fairly popularly know as the following

1) Time to reach maximum height, t, is half of the total time of flight T

t = (u sin θ)/g = (30 × sin 30°)/9.8 = 1.531s

2) Maximum height reached, H

H = (u² sin² θ)/2g = (30² sin² 30°)/(2 × 9.8) = 11.48m

3) Time when the projectile hits the ground is the total time of flight, T,

T = 2t = (2u sin θ)/g = 3.062s

4) The range of the projectile is the horizontal distance it covers.

R = (u² sin 2θ)/g = (30² sin 60°)/9.8 = 79.53m

5) Speed of the projectile as it hits the ground, Vf

Vf = √((Vᵧ)² + Vₓ²)

The final velocity component in the y-direction, Vᵧ is calculated from the equations of motion, using the vertical component of the initial velocity for initial velocity.

Uᵧ = u sin θ = 30 sin 30° = 15 m/s

Vᵧ = 15 - (9.8×3.062) = -15.076 m/s

The horizontal component of the final velocity, Vₓ

Vₓ = u cos θ = 30 cos 30 = 25.98 m/s

Vf = √((-15.076)² + (25.98)²) = 30 m/s.

Hope this Helps!!!

An image of the variation in velocity as the projectile moves is attached to this solution.

x=-2

infinite number of lines, so there is an infinite number of equations.

step-by-step explanation:

there are an infinite number of line that passes through the point (-2,5)

x=-2   is one line that passes through the point

y = 5 is another line that passes through the point

(these two lines are perpendicular to each other).

since the number of   numbers are infinite and the slope is the ratio of two numbers,(change in y over change in x) we can have an infinite number of slopes.     our only condition is that it passes through the point (-2,5)

step-by-step explanation:

we are given a graph for system of inequalities.

in order to write the system of inequalities for the given graph, we need to find the inequality for each of the line separately.

let us find the slope of dotted line first.

taking two coordinates on the graph.

let us say (0,0) and (3,1).

let us find the slope of the line now.

slope of the dotted line would be =.

y-intercept of dotted line is y=0 only.

plugging slope m= and b=0 in slope-intercept form y=mx+b, we get

.

the line is shaded up of that dotted line.

therefore, for shading up we would take greater than inequality symbol > .

and because it's a dotted line we would take only > symbol in place of = sign in above equation of line.

therefore, inequality for dotted line would be

let us take two points (3,1) and (0,4) on the solid line.

slope of the solid line would be =.

for the solid line we have y-intercept b= 4.

plugging slope m= -1 and b=4 in slope-intercept form y=mx+b, we get

y = -1x + 4.

the line is shaded down of that solid line.

therefore, for shading down we would take less than inequality symbol < .

and because it's a solid line we would take ≤ symbol in place of = sign in above equation of line.

therefore, inequality for solid line would be