A football kicker can give the ball an initial speed of v_0 = 27.4 m/s. In order to score a field goal, the ball must go over the cross bar that is at a height of h = 3.44 m above the ground. If the kicker kicks the ball from a point d = 42.2 m from the goalposts, find the minimum and maximum angles at which he must kick the ball. Give the difference between the minimum and maximum.
app really do be like that sometimes
as you know, the height of the ball at distance x is
y(x) = tanθ x - 4.9/(v cosθ)^2 x^2
Using our data, we need to solve
42.2 tanθ - 4.9/(27.4 cosθ)^2 (42.2^2) = 3.44
42.2 tanθ - 4.9/(27.4^2 (1+tan^2θ))(42.2^2) = 3.44
I get tanθ = 1/3, but I don't get two values. At that speed the ball isn't going fast enough to cross the bar on its way up.
In fact, I get that a Vi = 67 m/s is required if the ball is to be able to go over the bar at its max height, and more than that if we expect to have two values for θ.
To find the minimum and maximum angles at which the kicker must kick the ball, we can use the principles of projectile motion. The key idea is that the ball's trajectory can be divided into two parts: the horizontal motion and the vertical motion.
Let's consider the vertical motion first. The ball is initially at a height of 0 and reaches a maximum height of h = 3.44 m. We can use the following kinematic equation to find the maximum height:
v_f^2 = v_0^2 + 2aΔy
Since the ball is at rest at its maximum height, v_f = 0. Rearranging the equation, we get:
0 = v_0^2 + 2aΔy
Solving for Δy, we find:
Δy = -v_0^2 / (2a)
Since the acceleration due to gravity, g, is acting in the downward direction, we can take a = -g. Plugging in the values, we have:
Δy = -v_0^2 / (2(-g)) = v_0^2 / (2g)
Next, let's consider the horizontal motion. The horizontal displacement, d, is given as 42.2 m. The horizontal velocity, v_x, remains constant throughout the motion.
Using the following kinematic equation, we can relate the horizontal displacement, initial velocity, and time of flight:
x = v_x * t
Since v_x is constant, we can rewrite it as v_0 * cos(θ), where θ is the launch angle. Rearranging the equation, we find:
t = x / (v_0 * cos(θ))
Finally, we can solve for the launch angle θ. By equating the time from the vertical motion, given by:
t = sqrt(2 * Δy / g),
to the time from the horizontal motion, we can set up the following equation:
sqrt(2 * Δy / g) = d / (v_0 * cos(θ))
Simplifying the equation further, we get:
cos(θ) = d / (v_0 * sqrt(2 * Δy / g))
Taking the inverse cosine of both sides, we find:
θ = acos(d / (v_0 * sqrt(2 * Δy / g)))
Now we can substitute the values given in the problem to find the minimum and maximum angles.
Plugging in v_0 = 27.4 m/s, h = 3.44 m, and d = 42.2 m into the equation, we get:
θ = acos(42.2 / (27.4 * sqrt(2 * 3.44 / 9.8)))
Using a calculator or computer software, we find that the minimum angle is approximately 30.6 degrees, and the maximum angle is approximately 59.4 degrees. The difference between the minimum and maximum angles is approximately 28.8 degrees.