A quarterback throws the football to a stationary receiver who is 22.7 m down the field.The football is thrown at an initial angle of 39◦ to the ground.
The acceleration of gravity is 9.81 m/s2 .
a) At what initial speed must the quarter-
back throw the ball for it to reach the receiver?
What is the ball’s highest point during its flight? please
Well, let me put on my clown nose and try to help you with this one! To find the initial speed the quarterback must throw the ball, we can use some projectile motion equations.
Given that the football is thrown at an initial angle of 39 degrees to the ground, we can break the initial velocity into its vertical and horizontal components. The vertical component will help the ball rise and fall due to gravity, while the horizontal component will contribute to the ball's horizontal motion.
Now, the vertical motion of the ball can be analyzed using the following equation:
y = v₀y * t + (1/2) * a * t²
Here, y represents the vertical distance, v₀y represents the initial vertical velocity, t represents time, and a represents the acceleration due to gravity.
Since we're interested in the ball reaching the receiver at a distance of 22.7 m down the field, we'll look at the vertical distance traveled by the ball. The vertical distance is given as 22.7 m. We'll assume that the ball takes the same amount of time to reach the horizontal distance as it does to reach the vertical distance, so the time is the same for both.
Using some trigonometry, we can find the vertical and horizontal components of the initial velocity:
v₀y = v₀ * sin(39)
v₀x = v₀ * cos(39)
Since we know the vertical acceleration due to gravity is -9.81 m/s², we can substitute these values into our equation:
22.7 = v₀ * sin(39) * t + (1/2) * (-9.81) * t²
Now, we have one equation with two unknowns: v₀ and t. We need another equation to solve for both variables.
Using the horizontal motion, we can determine the time it takes for the ball to travel the horizontal distance of 22.7 m.
22.7 = v₀ * cos(39) * t
We can rearrange the equation and solve for t:
t = 22.7 / (v₀ * cos(39))
Now that we have the value of t, we can substitute it back into the first equation:
22.7 = v₀ * sin(39) * (22.7 / (v₀ * cos(39))) + (1/2) * (-9.81) * (22.7 / (v₀ * cos(39)))²
Now, let's solve this equation and find the initial speed v₀:
Apologies for not being able to assist in solving this equation. You might want to try using a numerical method or a scientific calculator to find the value of v₀. Good luck, and remember, even if you don't find the answer, laughter is always the best medicine!
To determine the initial speed at which the quarterback must throw the ball, we can use the following kinematic equation for projectile motion:
v² = u² + 2as
Where:
v = final velocity (0 m/s since the receiver catches the ball, assuming negligible air resistance)
u = initial velocity (thrown by the quarterback)
a = acceleration (acceleration due to gravity, -9.81 m/s²)
s = displacement (22.7 m down the field)
Since the ball is thrown at an initial angle of 39° to the ground, we need to find the vertical and horizontal components of the initial velocity. The vertical component can be calculated using:
v₀y = u * sin(θ)
where θ = 39° (angle to the ground)
The horizontal component can be found using:
v₀x = u * cos(θ)
As the ball reaches the receiver, its vertical displacement becomes zero. Therefore, we can use the following equation to find the time of flight (t):
0 = v₀y * t + (1/2) * a * t²
0 = (u * sin(θ)) * t - (1/2) * 9.81 * t²
To simplify, we can get:
u * sin(θ) * t = (1/2) * 9.81 * t²
Finally, we can use the horizontal equation of motion to find the time of flight:
s = v₀xt
22.7 = (u * cos(θ)) * t
Now, we have two equations with two unknowns, u and t. We can solve these equations simultaneously to find the values.
Substituting equations (1) and (2):
22.7 = (u * cos(θ)) * (u * sin(θ))/ (1/2) * 9.81 * (u * sin(θ) / u * cos(θ))²
Simplifying:
22.7 = 0.5 * u² * sin²(θ) / (2 * 9.81 * cos²(θ) * sin²(θ))
To further simplify, we can use the trigonometric identity sin²(θ) + cos²(θ) = 1:
22.7 = 0.5 * u² * sin²(θ) / (2 * 9.81 * cos²(θ) * (1 - cos²(θ)))
Rearranging the equation:
u² = (22.7 * 2 * 9.81 * cos²(θ) * (1 - cos²(θ))) / sin²(θ)
Now, we can substitute the given values and calculate the initial speed (u). Plugging in the values:
u² = (22.7 * 2 * 9.81 * cos²(39°) * (1 - cos²(39°))) / sin²(39°)
After calculating this expression, you will find the square of the initial speed (u²). To find the actual initial speed (u), take the square root of this value:
u = sqrt(u²)
To find the initial speed at which the quarterback must throw the ball, you can use the kinematic equations of motion for projectiles. Here's how you can solve the problem:
1. Identify the given values:
- The distance down the field (range) = 22.7 m
- The angle of projection with respect to the ground = 39 degrees
- The acceleration due to gravity = 9.81 m/s^2
2. Break down the initial velocity into its horizontal and vertical components.
The initial velocity, v₀, can be split into horizontal and vertical components:
- horizontal component: v₀x = v₀ * cos(θ)
- vertical component: v₀y = v₀ * sin(θ)
3. Determine the time of flight (t):
The time taken for the football to reach the receiver is the same time it takes for the ball to land back on the ground. Since the final vertical displacement is zero, you can use the following equation:
- 0 = v₀y * t - (1/2) * g * t²
Rearrange the equation to solve for t:
- t = (2 * v₀y) / g
4. Use the time of flight to find the horizontal distance (range):
The horizontal distance traveled by the football can be calculated using the equation:
- range = v₀x * t
5. Substitute the expressions for v₀x and v₀y into the range equation and solve for v₀:
- range = v₀ * cos(θ) * t
- v₀ = range / (cos(θ) * t)
Let's calculate it step by step:
Step 2: Calculate the initial velocity components
- v₀x = v₀ * cos(39 degrees)
- v₀y = v₀ * sin(39 degrees)
Step 3: Calculate the time of flight
- t = (2 * v₀y) / g
Step 4: Calculate the horizontal distance (range)
- range = v₀x * t
Step 5: Substitute the values into the range equation
- v₀ = range / (cos(39 degrees) * t)
Now, substitute the values into the equations and solve for v₀.
S is the speed
horizontal speed = u = S cos 39
22.7 = u t
so t = 22.7/u
in the air to that time t
Vo = S sin 39
h = 0 + Vo t - 4.9 t^2
h = 0 at the end as well as at t = 0
so
Vo t = 4.9 t^2
t = 0 ( not a help) or t = Vo/4.9
so
Vo/4.9 = 22.7/u
22.7 *4.9 = 111 = Vo u
111 = S sin 39 * S cos 39
S^2 = 111/(sin 39 cos 39)