A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity components are V0x = 310 m/s and V0y = 26 m/s. The projectile reaches maximum height at point P, then it falls and strikes the ground at point Q, which is 20 m below the launch point. For how long is the projectile in the air?

Yo = 26 m/s.=Ver. component of initial velocity.

Y = Yo + gt.
Tr = (Y-Yo)/g = (0-26/-9.8 = 2.65 s. =
Rise time.

Tf1 = Tr = 2.65 s. = Fall time to launching point.

h = Yo*t + 0.5g*t^2 = 20 m.
26t + 4.9t^2 = 20
4.9t^2 + 26t - 20 = 0
Use Quad. Formula:
Tf2 = 0.68 s. To fall 20 m.

T = Tr + Tf1 + Tf2=2.65 + 2.65 + 0.68 =
5.98 s. = Time in air.

4.5s

To find the time for which the projectile is in the air, we can use the equation of motion for vertical motion:

h = V0y * t - (1/2) * g * t^2,

where:
h is the height (relative to the ground),
V0y is the initial vertical velocity,
t is the time of flight, and
g is the acceleration due to gravity (which is approximately 9.8 m/s^2).

At the maximum height, the vertical velocity becomes zero. Therefore, we can find the time taken to reach the maximum height using the equation:

0 = V0y - g * t_max,

where t_max is the time taken to reach the maximum height.

Solving for t_max, we get:

t_max = V0y / g.

Given the initial vertical velocity V0y = 26 m/s, and the acceleration due to gravity g ≈ 9.8 m/s^2, we have:

t_max = 26 m/s / 9.8 m/s^2 = 2.653 seconds.

Since the projectile reaches maximum height at point P and falls down to the ground at point Q, the total time of flight is twice the time taken to reach the maximum height.

Therefore, the projectile is in the air for:

2 * t_max = 2 * 2.653 seconds = 5.306 seconds.

So, the projectile is in the air for approximately 5.306 seconds.

To find the time of flight of the projectile, we need to analyze its motion in the horizontal and vertical directions separately.

In the horizontal direction, the projectile moves with a constant velocity. Therefore, the time of flight, t, can be determined by considering the horizontal displacement, which is equal to the product of the horizontal velocity component, V₀ₓ, and the time of flight:

Horizontal displacement (x) = V₀ₓ * t

Since the projectile is fired from the origin (x = 0) and strikes the ground at point Q, the horizontal displacement is zero. Therefore, we can write the equation as:

0 = V₀ₓ * t

Solving for t, we find:
t = 0 / V₀ₓ

Since the horizontal velocity component, V₀ₓ, is 310 m/s, dividing zero by 310 will result in t being equal to zero seconds.

Now, let's analyze the vertical motion of the projectile. We know that the projectile reaches its maximum height at point P and then falls to the ground at point Q. The total vertical displacement can be calculated by subtracting the initial height of the projectile (y = 0) from the final height at point Q, which is -20 m:

Vertical displacement (y) = -20 m - 0 m
= -20 m

Using the equations of motion, we can determine the time it takes for the projectile to reach the maximum height by using the vertical velocity component, V₀ᵧ, and the acceleration due to gravity, g:

Vertical displacement (y) = V₀ᵧ * t + (1/2) * g * t²

Since the projectile reaches its maximum height at point P, the vertical displacement is zero:

0 = V₀ᵧ * t + (1/2) * g * t²

To solve this equation, we can use the quadratic formula:

t = (-b ± √(b² - 4ac)) / 2a

In this case, the equation is in the form of a quadratic equation:

(1/2) * g * t² + V₀ᵧ * t = 0

Using the values of the variables:
a = (1/2) * g
b = V₀ᵧ
c = 0

Substituting these values into the quadratic formula:

t = (-V₀ᵧ ± √(V₀ᵧ² - 4 * (1/2) * g * 0)) / 2 * (1/2) * g

Simplifying the equation:

t = (-V₀ᵧ ± √(V₀ᵧ²)) / g

Since time cannot be negative, we take the positive root of the equation:

t = (-V₀ᵧ + √(V₀ᵧ²)) / g

Given that V₀ᵧ is 26 m/s and g is approximately 9.8 m/s², we can calculate:

t = (-26 + √(26²)) / 9.8

By evaluating this expression, we find that the time it takes for the projectile to reach the maximum height is approximately 2.65 seconds.

The total time of flight is equal to twice the time it takes to reach the maximum height since the projectile takes the same amount of time to descend to the ground:

Total time of flight = 2 * time to reach maximum height
= 2 * 2.65 s
≈ 5.3 seconds

Therefore, the projectile is in the air for approximately 5.3 seconds.