To determine the initial speed and angle at which the ball was kicked, we can use the equations of projectile motion.
First, let's break down the motion into horizontal and vertical components.
In the vertical direction, the ball experiences free fall due to gravity. We can use the equation:
y = v₀y * t - (1/2) * g * t²
Where:
y = vertical displacement (0 since the ball returns to the ground)
v₀y = initial vertical velocity (unknown)
g = acceleration due to gravity (approximately 9.8 m/s²)
t = time (3.2 seconds)
0 = v₀y * 3.2 - (1/2) * 9.8 * (3.2)²
Simplifying the equation:
0 = v₀y * 3.2 - 15.68
v₀y * 3.2 = 15.68
v₀y = 15.68 / 3.2
v₀y ≈ 4.9 m/s (rounded to one decimal place)
In the horizontal direction, the ball moves with constant velocity. The equation to calculate horizontal distance is:
x = v₀x * t
Where:
x = horizontal range (29 m)
v₀x = initial horizontal velocity (unknown)
t = time (3.2 seconds)
29 = v₀x * 3.2
v₀x = 29 / 3.2
v₀x ≈ 9.06 m/s (rounded to two decimal places)
Now, to find the initial speed, we can use the Pythagorean theorem:
v₀ = √(v₀x² + v₀y²)
v₀ = √((9.06)² + (4.9)²)
v₀ ≈ √(82.0836 + 24.01)
v₀ ≈ √106.0936
v₀ ≈ 10.3 m/s (rounded to one decimal place)
Lastly, to find the angle at which the ball was kicked, we can use the equation:
θ = tan^(-1)(v₀y / v₀x)
θ = tan^(-1)(4.9 / 9.06)
θ ≈ tan^(-1)(0.5413)
θ ≈ 29.7° (rounded to one decimal place)
Therefore, the initial speed at which the ball was kicked is approximately 10.3 m/s and the angle is approximately 29.7 degrees.
during the opening kickoff of a college football game, the kicker makes a mistake and kicks the ball too high. the ball stays in the air for 3.2 seconds and has a horizontal range of only 29m. determine the initial speed and angle at which the ball was kicked.
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