A placekicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 24.0 m/s at an angle of 51.0° to the horizontal.
(a) By how much does the ball clear or fall short of clearing the crossbar? (Enter a negative answer if it falls short.)
m
(b) Does the ball approach the crossbar while still rising or while falling?
rising
falling
u = horizontal speed forever = 24 cos 51 = 15.1 m/s
v = vertical speed = 24 sin 51 - 9.81 t = 18.7 - 9.81 t
x = horizontal distance = 15.1 t
y = height = 18.7 t - 4.9 t^2
so
a)
time at crossbar = 36.0 / 15.1 = 2.38 seconds
height at crossbar = 18.7 (2.38) - 4.9 (2.38)^2 = 44.5 - 27.8 = 16.7 meters
so 16.7 - 3.05
b) when is the vertical speed = 0?
v= 18.7 -9.81 t
so t at top = 1.9 seconds
but it takes 2.38 seconds to reach crossbar
so it is on the way down
To solve this problem, we can use the equations of projectile motion. Let's break down the problem step by step.
Step 1: Determine the time it takes for the ball to reach its highest point.
We know that the vertical motion of the ball is governed by the equation:
h = v₀y * t - (1/2) * g * t²
where:
- h is the vertical displacement (height)
- v₀y is the vertical component of the initial velocity
- t is the time
- g is the acceleration due to gravity (approximately 9.8 m/s²)
At the highest point of the ball's trajectory, the vertical displacement h will be the difference between the height of the crossbar (3.05 m) and the initial height (0 m).
3.05 m = v₀y * t - (1/2) * 9.8 m/s² * t²
Step 2: Solve the equation for time.
Rearranging the equation, we have:
(1/2) * 9.8 m/s² * t² + v₀y * t - 3.05 m = 0
This is a quadratic equation in terms of t. We can solve it using the quadratic formula:
t = (-b ± sqrt(b² - 4ac)) / 2a
where:
- a = 4.9 m/s² (1/2 * 9.8 m/s²)
- b = v₀y
- c = -3.05 m
Using the given initial velocity of 24.0 m/s and the launch angle of 51.0°, we can calculate the vertical component of the initial velocity:
v₀y = v₀ * sin(θ)
= 24.0 m/s * sin(51.0°)
Step 3: Calculate the time it takes for the ball to reach its highest point.
Substituting the values into the quadratic formula, we can find the two possible values for t:
t₁ = (-v₀y + sqrt(v₀y² - 4*a*c)) / (2*a)
t₂ = (-v₀y - sqrt(v₀y² - 4*a*c)) / (2*a)
However, in this case, we are only interested in the positive value because time cannot be negative:
t = t₁
Step 4: Calculate the maximum height reached by the ball.
Using the time t, we can substitute it into the equation for vertical motion to find the maximum height:
h_max = v₀y * t - (1/2) * g * t²
Step 5: Determine the vertical distance the ball clears or falls short of the crossbar.
The vertical distance the ball clears or falls short of the crossbar is the difference between the maximum height and the height of the crossbar:
distance = h_max - 3.05 m
This will give us the answer for question (a).
To answer question (b), if the distance is positive (greater than 0), this means the ball cleared the crossbar and approached it while falling. If the distance is negative (less than 0), the ball fell short of the crossbar and approached it while still rising.
Now you can solve the problem using the explained steps and find the answers to (a) and (b).