A football kicker is trying to make a field goal from 42.0m. If he kicks at 35∘ above the horizontal, what minimum speed is needed to clear the 3.10m high crossbar?
Equation of trajectory:
y = (g x^2 sec^2(θ))/(2 v^2) + x tan(θ)
So, given:
θ = 35°
x = 42.0
y = 3.10
Find v
To find the minimum speed needed to clear the crossbar, we can use principles of projectile motion. Here's how you can calculate the minimum speed:
Step 1: Identify the given information:
- Initial horizontal distance (range), R = 42.0 m
- Launch angle, θ = 35° above the horizontal
- Vertical displacement (height), h = 3.10 m
Step 2: Analyze the vertical motion:
The football needs to clear the crossbar, which means it should reach a maximum height greater than or equal to the height of the crossbar. To find the maximum height, we can use the vertical motion equation:
h = (v₀y²) / (2g)
Where:
- v₀y is the initial vertical velocity
- g is the acceleration due to gravity (9.8 m/s²)
Step 3: Find the initial vertical velocity:
To find the initial vertical velocity, we can use the launch angle:
v₀y = v₀ * sin(θ)
Where:
- v₀ is the initial velocity
Step 4: Analyze the horizontal motion:
The horizontal distance traveled can be calculated using the range formula:
R = (v₀x * t)
Where:
- v₀x is the initial horizontal velocity
- t is the time of flight
Since there is no vertical acceleration in horizontal motion, v₀x remains constant throughout the flight.
Step 5: Find the initial horizontal velocity:
To find the initial horizontal velocity, we can use the launch angle:
v₀x = v₀ * cos(θ)
Step 6: Calculate the time of flight:
The time of flight is the total time taken for the football to reach the crossbar. We can calculate it using the vertical motion equation:
h = (v₀y * t) - (0.5 * g * t²)
Step 7: Solve for v₀ (initial velocity):
Using the horizontal motion equation, we can express v₀x in terms of t and R:
v₀x = R / t
Substituting the values of v₀x and v₀y into the equation, we can solve for v₀:
v₀ = √((v₀x)² + (v₀y)²)
Step 8: Substitute all the given values into the equation and solve for v₀.
Vy(f)^2=Vy(i)^^+2gd
Where
Vy(f)=0m/s
Vy(i)=?
g=9.8m/s^2
d=3.10
Solve for Vy(i)
0=Vy(i)^2+2(-9.8)(3.10m)
Vy(i)^2=60.8m
Vy(i)=sqrt*[60.8m]
Vy(i)=7.79m/s
This is the velocity in the y-direction:
V(i)*Sin35º=7.79m/s
Solve for V(i):
V(i)=7.79m/s/Sin35º
V(i)=13.6m/s