A football is kicked at ground level with a speed of 19.8 m/s at an angle of 32.2 degrees to the horizontal.

Question: How much later does the football hit the ground?

t=2•vₒ•sinα/g

2.15 seconds

The Builder

To determine how much later the football hits the ground, we need to find the time it takes for the football to reach the ground after being kicked.

We can break down the initial velocity of the football into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity.

Given:
Initial speed (v) = 19.8 m/s
Launch angle (θ) = 32.2 degrees

First, let's find the vertical component of the initial velocity (v_y):
v_y = v * sin(θ)

v_y = 19.8 m/s * sin(32.2°)
v_y = 10.5 m/s

Now, let's find the time the football stays in the air. We'll use the equation of motion for vertical motion:
y = v_y * t - 0.5 * g * t^2

Since the football is kicked at ground level, the displacement in the vertical direction (y) is zero when it hits the ground. We can set y = 0 in the equation above and solve for t:

0 = (10.5 m/s) * t - 0.5 * g * t^2

Now, we need to know the acceleration due to gravity (g), which is approximately 9.8 m/s^2.

0 = (10.5 m/s) * t - 0.5 * (9.8 m/s^2) * t^2

Rearranging the equation to find t:
0.5 * (9.8 m/s^2) * t^2 - (10.5 m/s) * t = 0

Using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a), where a = 0.5 * (9.8 m/s^2), b = -10.5 m/s, and c = 0.

t = (-(-10.5) ± √((-10.5)^2 - 4 * (0.5 * (9.8)) * 0)) / (2 * 0.5 * (9.8))

Simplifying the equation, we get:
t = (10.5 ± √(110.25)) / 4.9

Now we solve for t using the positive square root value:
t = (10.5 + √(110.25)) / 4.9
t ≈ 3.22 seconds

Therefore, the football hits the ground approximately 3.22 seconds after being kicked.