A punter kicks a football at an angle of 45 degrees to the ground. The football has an initial velocity of 25 m/s, make a diagram and calculate:
a) How long was the football in the air?
b) How far does the football travel horizontally?
c) What is the maximum height of the football?
a player kicks a football at an angle of 40 degrees with an initial speed of 48 ft/second. a second player standing at a distance of 100 ft from the first in the direction of the kick starts running to meet the ball at the instant it is kicked. how fast must he run in order to catch the ball before it hits the ground
To determine the motion of the football, we can break down the initial velocity into its horizontal and vertical components. Let's create a diagram to visualize the situation.
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The longer arrow on the left represents the initial velocity vector (25 m/s) of the football. The vertical arrow (vertical component) is labeled as "vt," and the horizontal arrow (horizontal component) is labeled as "vh."
a) To calculate the time of flight, we need to find the time it takes for the football to reach its highest point and then return to the ground. The vertical motion is affected by gravity, but the horizontal motion is not. The vertical component of the velocity (vt) will decrease until reaching zero when the ball is at its maximum height.
Using the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time, we can find the time it takes for vt to reach zero:
0 = vt - g * t
Solving for t:
vt = g * t
t = vt / g
Since the vertical motion involves two identical halves (up and down) with equal time, we can multiply the result by 2:
t_total = 2 * t
The acceleration due to gravity is approximately 9.8 m/s^2. Let's substitute the given values:
t = vt / g = (25 m/s * sin 45°) / 9.8 m/s^2 = (25 * 0.707) / 9.8 ≈ 1.79 s
t_total = 2 * t ≈ 2 * 1.79 ≈ 3.58 s
Therefore, the football spends approximately 3.58 seconds in the air.
b) To calculate the horizontal distance traveled by the football, we apply the equation d = vh * t:
vh = 25 m/s * cos 45° ≈ 25 * 0.707 ≈ 17.68 m/s
d = vh * t ≈ 17.68 m/s * 3.58 s ≈ 63.27 m
Hence, the football travels approximately 63.27 meters horizontally.
c) The maximum height of the football occurs at the midpoint of its flight, where its vertical velocity becomes zero.
Using the equation vf = vi + at, we can determine this height. Since the initial vertical velocity (vi) is equal to the final vertical velocity (vf) at the maximum height, we have:
0 = vt - g * t_max
Solving for t_max:
vt = g * t_max
t_max = vt / g
t_max = (25 m/s * sin 45°) / 9.8 m/s^2 ≈ (25 * 0.707) / 9.8 ≈ 1.79 s
Next, we use the equation d = vi * t + 0.5 * a * t^2 to find the displacement in the vertical direction:
d_max = vt * t_max + 0.5 * (-9.8 m/s^2) * t_max^2
d_max ≈ (25 * 0.707) * 1.79 + 0.5 * (-9.8) * (1.79)^2
d_max ≈ 17.68 * 1.79 - 0.5 * 9.8 * 1.79^2
d_max ≈ 31.66 m - 15.37 m
d_max ≈ 16.29 m
Therefore, the maximum height of the football is approximately 16.29 meters.
To calculate the answers, we will break down the motion of the football into its horizontal and vertical components.
First, let's work on the horizontal component. Since there is no acceleration in the horizontal direction, the initial velocity will remain constant throughout the motion.
a) To find the time the football is in the air, we can use the vertical component. We can split the initial velocity (25 m/s) into its vertical and horizontal components using trigonometry. The vertical component is given by V_y = V * sin(θ).
V_y = 25 * sin(45°) = 17.68 m/s
The time of flight can be calculated using the equation t = 2 * V_y / g, where g is the acceleration due to gravity (9.8 m/s^2).
t = 2 * 17.68 / 9.8 = 3.2 seconds (approximately)
b) To find the horizontal distance traveled, we can use the horizontal component of the initial velocity. The horizontal component is given by V_x = V * cos(θ).
V_x = 25 * cos(45°) = 17.68 m/s
The horizontal distance can be calculated using the equation d = V_x * t.
d = 17.68 * 3.2 = 56.38 meters (approximately)
c) To find the maximum height, we need to look at the vertical motion. At the highest point of the trajectory, the vertical velocity will be zero. We can use this information to find the time taken to reach the maximum height, and then calculate the height using the equation h = V_y^2 / (2 * g).
At the highest point, V_y = 0.
V_y = V * sin(θ)
0 = 25 * sin(45°)
sin(45°) = 0 (This is incorrect, sin(45°) = √2 / 2)
To correct this, let's find the correct angle.
sin(θ) = V_y / V
θ = sin^(-1)(V_y / V)
θ = sin^(-1)(17.68 / 25) ≈ 40.2°
Now, we can use the corrected angle to find the maximum height.
h = (V_y^2) / (2 * g)
h = (17.68^2) / (2 * 9.8) ≈ 16.70 meters
To summarize:
a) The football was in the air for approximately 3.2 seconds.
b) The football traveled horizontally for approximately 56.38 meters.
c) The maximum height of the football was approximately 16.70 meters.
viy=25sin45
vix=25cos45
a)let height be zero (y=0)
you can use y=vit+1/2at2, the use quadratic formula to find t
b)after you find t in part a...use x=vixt
x will be the horizontal distance of the ball
c) vfy=0m/s
use vfy^2=vi^2+2ay
y will be the maximum height
a=g=9.8m/s^2
..
bye...