A quarterback throws the football to a stationary receiver who is 18.3 m down the field. Football is thrown at an angle of 44.9 degrees to the ground. Acceleration is 9.81 m/s2. What initial spped must the quarterback throw ball to reach receiver and what is the highest point during it's flight.
Split the problem in two, rising and falling. It does half the time and distance rising. (parabolic path)
Initial speed = S
U = constant speed in x direction = S cos 44.9 = .708 s
Vo = initial speed in y direction = S sin 44.9 = .706 s
vertical speed at max height = 0
so
0 = Vo - 9.81 t
where t is time rising (half of total time in the air)
t = .706 s/9.81 = .0720 s
In half the time in the air (t) it goes half the distance down field , 9.15 m
U t = 9.15
t = 9.15 /(.708 s)
so
.072 s = 9.15 / (.708 s)
.051 s^2 = 9.15
s^2 = 180
s = 13.4 m/s
now how high?
t = 9.15/(.708 s) = 9.15/(.708*13.4) = .964 seconds rising
h = Vo t - 4.9 t^2
h = .706*13.4 *.964 - 4.9 (.964^2)
= 4.56 meters (about 14 feet)
The horizontal velocity is Vi*cos44.9 and the vertical velocity is Vi*sin44.9
horizontal
18.3=Vi*cos44.9 * time find time in terms of Vi.
Vertical
hf=hi+Vi*sin44.9*t -1/2 g t^2
now hf=hi=0
so
0=Vi*sin44.9 -1/2 g t and put the expression you got for t, then solve for Vi. You will encounter the need for one trig identity.
To find the initial speed at which the quarterback must throw the ball and the highest point during its flight, we can use the principles of projectile motion.
Let's break down the question into two parts - finding the initial speed and finding the highest point.
Part 1: Finding the initial speed:
In projectile motion, the horizontal and vertical motions are independent of each other.
Given information:
- Distance traveled horizontally (x) = 18.3 m
- Angle of the throw (θ) = 44.9 degrees
- Acceleration due to gravity (g) = 9.81 m/s^2
To find the initial speed (v0), we need to analyze the horizontal and vertical components of the velocity.
The horizontal component of the initial velocity remains constant throughout the motion. We can find it using the equation:
Horizontal component: vx = v0 * cos(θ)
The vertical component of the initial velocity changes due to gravity. We can find it using the equation:
Vertical component: vy = v0 * sin(θ) - g * t
At the highest point of the motion, the vertical component of the velocity becomes zero since the object momentarily stops going up before falling back down.
To find the time taken to reach the highest point (t), we can use the equation:
vy = v0 * sin(θ) - g * t = 0
t = v0 * sin(θ) / g
Using this time value, we substitute it into the previous equation to find the initial speed:
0 = v0 * sin(θ) - g * (v0 * sin(θ) / g)
0 = v0 * sin(θ) - v0 * sin(θ)
0 = 0
Since the equation results in 0, we can't solve for v0 using this method. However, we can find the range of initial speeds that will allow the ball to reach the receiver.
Instead of using the highest point, let's use the time of flight (total time taken to reach the receiver at 18.3m). The time of flight (T) is given by the equation:
T = 2 * t = 2 * v0 * sin(θ) / g
Rearranging the equation to solve for v0:
v0 = T * g / (2 * sin(θ))
Given:
- Distance traveled horizontally (x) = 18.3 m
- Angle of the throw (θ) = 44.9 degrees
- Acceleration due to gravity (g) = 9.81 m/s^2
Calculating:
v0 = (2 * 18.3 * 9.81) / (2 * sin(44.9))
v0 ≈ 20.09 m/s
Therefore, the initial speed at which the quarterback must throw the ball is approximately 20.09 m/s.
Part 2: Finding the highest point:
We can find the highest point by analyzing the vertical motion.
Using the equation for vertical position:
Vertical component: y = v0 * sin(θ) * t - (1/2) * g * t^2
At the highest point, the vertical velocity component is zero. Therefore:
0 = v0 * sin(θ) - g * t
t = v0 * sin(θ) / g
Using this time value, we can substitute it back into the equation for vertical position to find the height (y) at the highest point:
y = v0 * sin(θ) * (v0 * sin(θ) / g) - (1/2) * g * (v0 * sin(θ) / g)^2
y = v0^2 * sin^2(θ) / (2 * g)
Substituting the given values:
y = (20.09^2 * sin^2(44.9)) / (2 * 9.81)
y ≈ 7.87 m
The highest point during the ball's flight is approximately 7.87 meters.
To find the initial speed at which the quarterback must throw the ball, we can use the kinematic equation for the horizontal distance traveled:
d = v₀ * t
where d is the horizontal distance (18.3 m) and t is the time it takes for the ball to reach the receiver.
To find the time, we need to break down the initial velocity of the ball into its horizontal and vertical components. We can use the equation:
v₀x = v₀ * cosθ
v₀y = v₀ * sinθ
where v₀x is the horizontal component of the initial velocity, v₀y is the vertical component of the initial velocity, and θ is the angle at which the ball was thrown (44.9 degrees).
The horizontal distance traveled can be found using the equation:
d = v₀x * t
where t is the time it takes for the ball to reach the receiver.
From the equations above, we can substitute the values to get:
18.3 m = (v₀ * cos44.9) * t
Since we have two unknowns (v₀ and t) and only one equation, we need another equation to solve for both variables.
Next, let's consider the vertical motion of the ball.
The equation for the vertical distance traveled is given by:
y = v₀y * t + (1/2) * a * t²
where y is the vertical distance traveled (ending position - starting position), v₀y is the vertical component of the initial velocity, a is the acceleration due to gravity (-9.81 m/s²), and t is the time.
Since the ball reaches its highest point at the midpoint of its flight, the vertical distance traveled at this point is 0. Therefore, we can set y = 0. Substituting the values:
0 = (v₀ * sin44.9) * t + (1/2) * (-9.81) * t²
Now, we have two equations:
18.3 = (v₀ * cos44.9) * t (Equation 1)
0 = (v₀ * sin44.9) * t + (1/2) * (-9.81) * t² (Equation 2)
We can solve these equations simultaneously to find the values of v₀ and t.
Once we have the value of t, we can substitute it back into Equation 1 to find the initial speed (v₀).
To find the highest point during the ball's flight, we need to calculate the time it takes for the ball to reach that point. At the highest point, the vertical velocity of the ball is momentarily zero. We can find this time using the equation:
v = v₀y + a * t
Setting v = 0, we can solve for t:
0 = (v₀ * sin44.9) + (-9.81) * t
Solving for t will give us the time it takes for the ball to reach its highest point. We can then substitute this value into the equation for the vertical distance traveled to find the height at the highest point.