A football is punted and leaves the punter’s foot at a height of 1.00 m above the ground. The football’s initial velocity is 21.5 m/s 35o above the horizontal.
(a) How far did the football travel horizontally before hitting the ground?
(b) What is the football’s velocity upon contact with the ground? Specify magnitude and direction
the height is
h = 1 + (21.5 sin35°)t - 4.9t^2
h=0 when t=2.6
the horizontal speed is a constant 21.5 cos35° m/s
so how far did it go in 2.6 seconds
the vertical speed is 21.5 sin35° - 9.8t
so now you can figure the velocity on impact
To find the horizontal distance the football traveled before hitting the ground, we can use the equation for horizontal motion:
d = vt
where:
d = horizontal distance
v = horizontal velocity
t = time
In this case, we need to find the horizontal component of the initial velocity. We can do this by using trigonometry:
v_horizontal = v * cosθ
where:
v = initial velocity
θ = angle with the horizontal
Using the given values:
v = 21.5 m/s
θ = 35 degrees
Calculating the horizontal component of the initial velocity:
v_horizontal = 21.5 m/s * cos(35 degrees)
v_horizontal ≈ 17.58 m/s
Now, we can find the time taken by the football to reach the ground. Since we know that the vertical motion is affected by gravity, we can use the equation:
h = ut + (1/2)gt^2
where:
h = initial height = 1.00 m
u = initial vertical velocity = v_vertical = v * sinθ = 21.5 m/s * sin(35 degrees)
g = acceleration due to gravity = 9.8 m/s^2 (assuming no air resistance)
t = time
Calculating the vertical component of the initial velocity:
v_vertical = 21.5 m/s * sin(35 degrees)
v_vertical ≈ 12.35 m/s
Now, we can solve for time:
1.00 m = (12.35 m/s)t - (1/2)(9.8 m/s^2)t^2
Rearranging the equation, we get a quadratic equation:
4.9t^2 - 12.35t + 1.00 = 0
We can solve this equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 4.9, b = -12.35, and c = 1.00.
Solving for t, we get:
t ≈ 2.1 s
Now, we can calculate the horizontal distance using the equation d = vt:
d = 17.58 m/s * 2.1 s
d ≈ 36.9 m
Therefore, the football traveled approximately 36.9 meters horizontally before hitting the ground.
To find the football's velocity upon contact with the ground, we need to find the vertical component of its velocity at that point. Since the football is experiencing free-fall, the vertical velocity at the point of impact would be equal to the negative of the initial vertical velocity.
v_vertical_impact = -v_vertical ≈ -12.35 m/s
To find the magnitude of the football's velocity upon contact with the ground, we can use the Pythagorean theorem:
v_impact = √(v_horizontal^2 + v_vertical_impact^2)
v_impact = √(17.58^2 + (-12.35)^2)
v_impact ≈ 21.71 m/s
The direction of the football's velocity upon contact with the ground would be the opposite of the initial angle with the horizontal:
θ_impact = -θ = -35 degrees
Therefore, the magnitude of the football's velocity upon contact with the ground is approximately 21.71 m/s, and the direction is approximately 35 degrees below the horizontal.
To solve this problem, we can use the kinematic equations of motion. Let's first break down the given information:
Initial height (h) = 1.00 m
Initial velocity (v₀) = 21.5 m/s
Launch angle (θ) = 35°
To answer part (a), we need to find the horizontal distance traveled by the football. We can use the horizontal component of the initial velocity to calculate this.
Horizontal velocity component (v₀x) = v₀ * cos(θ)
v₀x = 21.5 m/s * cos(35°)
To find the horizontal distance (x) covered, we can use the equation:
x = v₀x * t
However, we don't have the time (t). To find the time, we can use the vertical motion of the football.
The vertical velocity component (v₀y) can be calculated using:
v₀y = v₀ * sin(θ)
v₀y = 21.5 m/s * sin(35°)
Since the ball starts at a height of 1.00 m and returns to the ground, we can use the equation:
h = v₀y * t + (1/2) * g * t²
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the values, we get:
1.00 m = (21.5 m/s * sin(35°)) * t + (1/2) * (9.8 m/s²) * t²
Solving this quadratic equation for t will give us the time it takes for the ball to hit the ground.
Once we have the value of t, we can substitute it back into the equation x = v₀x * t to find the horizontal distance covered by the football.
To answer part (b), we need to find the final velocity of the football just before it hits the ground. We can use the equation:
v = v₀y + g * t
where v is the final velocity.
The magnitude of the velocity is given by the equation:
|v| = √(v_x² + v_y²)
where v_x is the horizontal component of the velocity and v_y is the vertical component of the velocity. The direction of the velocity can be determined using trigonometry.
By following these steps, we can calculate the answers to parts (a) and (b) of the problem.