A football kicker can give the ball an initial speed of 30 m/s. What are the (a) least and (b) greatest elevation angles at which he can kick the ball to score a field goal from a point 48 m in front of goalposts whose horizontal bar is 3.44 m above the ground?
To find the least and greatest elevation angles at which the football kicker can score a field goal, we can use the concept of projectile motion.
First, let's analyze the vertical motion of the ball. The vertical motion can be described using the equation:
h = ut + (1/2)gt^2
where:
h = vertical height or displacement
u = initial vertical velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Since we want the ball to go through the goalposts, the vertical displacement (h) should be equal to the height of the horizontal bar of the goalpost, which is 3.44 m.
Substituting the values, we have:
3.44 = 0 + (1/2)(9.8)t^2
6.88 = 4.9t^2
Now let's analyze the horizontal motion of the ball. The range (R) of the projectile is given by:
R = u cosθ × t
where:
R = horizontal range or displacement
u = initial horizontal velocity
θ = angle of elevation
t = time
We know that the range of the projectile should be 48 m, as the ball needs to cover the distance to reach the goalpost.
Substituting the values, we have:
48 = 30 cosθ × t
Now we have two equations involving the time (t):
6.88 = 4.9t^2 (equation 1)
48 = 30 cosθ × t (equation 2)
We can now solve these equations simultaneously to find the values of t and θ.
To find the least elevation angle, we can substitute equation 2 into equation 1 and solve for t:
6.88 = 4.9t^2
t^2 = 6.88 / 4.9
t^2 = 1.404
t ≈ 1.18 seconds
Now, substituting t = 1.18 into equation 2 to find θ:
48 = 30 cosθ × 1.18
θ = arccos(48 / (30 × 1.18))
θ ≈ 57.2 degrees
So the least elevation angle at which the kicker can score a field goal is approximately 57.2 degrees.
To find the greatest elevation angle, we substitute equation 2 into equation 1 again:
6.88 = 4.9t^2
t^2 = 6.88 / 4.9
t^2 = 1.404
t ≈ 1.18 seconds
Now, substituting t = 1.18 into equation 2 to find θ:
48 = 30 cosθ × 1.18
θ = arccos(48 / (30 × 1.18))
θ ≈ 57.2 degrees
Therefore, both the least and greatest elevation angles at which the kicker can score a field goal are approximately 57.2 degrees.