A football is kicked at ground level with a speed of 17.9m/s at an angle of 39.0∘ to the horizontal. How much later does it hit the ground?

Vo = 17.9m/s[39o]

Xo = 17.9*cos39 = 13.91 m/s.
Yo = 17.9*sin39 = 11.26 m/s.

Y = Yo + g*Tr = 0 @ max ht.
11.26 - 9.8Tr = 0
9.8Tr = 11.26
Tr = 1.15 s. = Rise time.

Tf = Tr = 1.15 s. = Fall time.

Tr+Tf = 1.15 + 1.15 = 2.30 s. = Time to
hit gnd.

2.30s

To find out how much later the football hits the ground, we need to calculate the time it takes for the football to reach the ground.

First, let's break down the initial velocity of the football into its horizontal and vertical components.

Horizontal velocity (Vx) = V0 * cos(θ)
Vertical velocity (Vy) = V0 * sin(θ)

where:
V0 = initial speed = 17.9 m/s
θ = angle of projection = 39.0°

Now, let's calculate these components:

Vx = 17.9 m/s * cos(39.0°)
Vx ≈ 13.72 m/s

Vy = 17.9 m/s * sin(39.0°)
Vy ≈ 11.43 m/s

Next, we'll use the vertical component (Vy) to calculate the time it takes for the football to reach its maximum height.

Using the equation:
Vy = ay * t

where:
ay = acceleration due to gravity = -9.81 m/s² (negative because it's acting downwards)
t = time

Rearranging the equation, we have:
t = Vy / ay

t = 11.43 m/s / 9.81 m/s²
t ≈ 1.17 s

So, it takes approximately 1.17 seconds for the football to reach its maximum height.

Now, we'll use the time (t) to calculate the total time it takes for the football to hit the ground.

Since the flight path is symmetric, the total time would be twice the time it took to reach the maximum height.

Total time = 2 * t
Total time ≈ 2 * 1.17 s
Total time ≈ 2.34 s

Therefore, the football would hit the ground approximately 2.34 seconds later.

To find out how much later the football hits the ground, we need to determine the time it takes for the football to reach the ground after being kicked. We can use the vertical motion of the football to calculate this.

First, we need to separate the initial velocity into horizontal and vertical components. The horizontal component (Vx) remains constant throughout the motion, while the vertical component (Vy) changes due to the acceleration due to gravity.

Given:
Initial speed (V0) = 17.9 m/s
Angle (θ) = 39.0°

We can find the horizontal component (Vx) by using the cosine of the angle:
Vx = V0 * cos(θ)

Vx = 17.9 m/s * cos(39.0°)

Vx ≈ 13.7 m/s

The vertical component (Vy) can be found using the sine of the angle:
Vy = V0 * sin(θ)

Vy = 17.9 m/s * sin(39.0°)

Vy ≈ 11.0 m/s

Next, we can use the vertical motion equation to determine the time it takes for the football to reach the ground. The equation is:

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

Where:
h = vertical displacement (height of the football above the ground) = 0 (since it hits the ground)
g = acceleration due to gravity = 9.8 m/s^2
t = time

Plugging in the values, we get:

0 = 11.0 m/s * t - (1/2) * 9.8 m/s^2 * t^2

Rearranging the equation:

(1/2) * 9.8 m/s^2 * t^2 - 11.0 m/s * t = 0

We can solve this quadratic equation using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / (2a)

Where:
a = (1/2) * 9.8 m/s^2
b = -11.0 m/s
c = 0

Plugging in the values:

t = (-(-11.0) ± sqrt((-11.0)^2 - 4 * (1/2) * 9.8 * 0)) / (2 * (1/2) * 9.8)

Simplifying the equation:

t = (11.0 ± sqrt(121 - 0)) / (9.8)

t = (11.0 ± sqrt(121)) / (9.8)

t = (11.0 ± 11.0) / (9.8)

This gives us two possible values for time:
t1 ≈ 2.02 s
t2 ≈ 0.0 s (discard this since it represents the initial time)

Therefore, the football hits the ground approximately 2.02 seconds later.