A quarterback throws the football to a stationary receiver who is 22.7m down the field. the football is thrown at an initial angle of 39 degrees to the ground. The acceleration of gravity is 9.81m/s^2.
a) at what initial speed must the quarterback throw the ball for it to reach the receiver? Answer in units of m/s
b) What is the ball’s highest point during its
flight?
To solve this problem, we can use the equations of motion for projectile motion. We'll start by finding the initial speed at which the quarterback must throw the ball for it to reach the receiver (part a).
a) To find the initial speed, we can use the horizontal and vertical components of the motion separately.
We know that the horizontal velocity (Vx) remains constant throughout the motion, while the vertical velocity (Vy) changes due to the acceleration of gravity.
Since the receiver is stationary, the horizontal component of the motion won't be affected. Therefore, we can find the horizontal velocity using the equation:
Vx = V * cos(theta)
where V is the initial speed and theta is the launch angle (in degrees).
In this case, Vx = V * cos(39 degrees).
Now, let's find the vertical component of the motion. The vertical displacement (Sy) can be calculated using the equation:
Sy = Vyi * t + (1/2) * a * t^2
where Vyi is the initial vertical velocity, t is the time of flight, and a is the acceleration due to gravity.
Since the receiver is at the same height as the quarterback, we can say that the vertical displacement is zero (Sy = 0).
Substituting the values, we have:
0 = V * sin(theta) * t - (1/2) * g * t^2
Rearranging this equation, we get:
t = (2 * V * sin(theta)) / g
Now, let's substitute this value of t into our equation for Vx:
Sy = V * cos(theta) * t
But we know that Sy = 22.7m, so,
22.7m = V * cos(39 degrees) * ((2 * V * sin(39 degrees)) / g)
Simplifying, we get:
22.7m = (2 * V^2 * cos(39 degrees) * sin(39 degrees)) / g
Now, we can solve this equation to find the value of V, the initial speed.
b) To find the ball's highest point during its flight (part b), we need to find the time it takes to reach the highest point. This can be found using the equation:
Vy = Vyi - g * t
At the highest point, Vy will become zero because the velocity changes direction. Therefore, we can set Vy = 0 and solve for t.
0 = V * sin(theta) - g * t
Solving for t, we have:
t = (V * sin(theta)) / g
Now we have the time when the ball reaches its highest point. We can substitute this value of t into the equation for the vertical displacement (Sy) to calculate the highest point.
a) To find the initial speed at which the quarterback must throw the ball, we can use the equation of motion for projectile motion. The horizontal motion and vertical motion can be considered separately.
In the horizontal direction, the velocity remains constant throughout the motion, so we have:
Horizontal velocity (Vx) = initial speed (Vo) * cos(theta), where theta is the initial angle of 39 degrees.
Vx = Vo * cos(39)
In the vertical direction, the acceleration is due to gravity, and the displacement is -22.7m (taking downwards as negative). We can use the kinematic equation:
Displacement (Y) = Vo * sin(theta) * t + (1/2) * (-g) * t^2
At the highest point of the ball's flight, the vertical velocity (Vy) is zero. Therefore:
0 = Vo * sin(theta) - g * t, where g is the acceleration due to gravity (9.81 m/s^2).
We can solve these equations simultaneously to find the initial speed (Vo):
0 = Vo * sin(theta) - g * t
t = (Vo * sin(theta)) / g
Substituting this value of t into the equation for displacement, we have:
-22.7 = Vo * sin(theta) * ((Vo * sin(theta)) / g) + (1/2) * (-g) * ((Vo * sin(theta)) / g)^2
Simplifying:
-22.7 = (Vo^2 * sin^2(theta)) / g - (Vo^2 * sin^2(theta)) / (2g)
Now, substitute sin^2(theta) with 1 - cos^2(theta):
-22.7 = (Vo^2 * (1 - cos^2(theta))) / g - (Vo^2 * (1 - cos^2(theta))) / (2g)
Multiplying through by 2g:
-22.7 * 2g = Vo^2 * (1 - cos^2(theta)) - Vo^2 * (1 - cos^2(theta)) / 2
-45.4g = Vo^2 * (1 - cos^2(theta)) * (1 - 1/2)
-45.4g = Vo^2 * (1 - cos^2(theta)) * 1/2
Vo^2 = (-45.4g) / (1 - cos^2(theta)) * 2
Finally:
Vo = sqrt((-45.4g) / (1 - cos^2(theta)) * 2)
Substituting the given values:
Vo = sqrt((-45.4 * 9.81) / (1 - cos^2(39)) * 2)
Vo ≈ 20.9 m/s
Therefore, the initial speed at which the quarterback must throw the ball for it to reach the receiver is approximately 20.9 m/s.
b) To find the ball's highest point during its flight, we can use the equation for vertical displacement:
Displacement in the vertical direction (Y) = Vo * sin(theta) * t + (1/2) * (-g) * t^2
At the highest point, the vertical displacement Y is maximum. Therefore, we can find the highest point by finding the time it takes for the ball to reach the highest point.
From the equation in part a, we have:
0 = Vo * sin(theta) - g * t
Solving for t:
t = (Vo * sin(theta)) / g
Substituting this value of t into the equation for displacement, we have:
Y = Vo * sin(theta) * ((Vo * sin(theta)) / g) + (1/2) * (-g) * ((Vo * sin(theta)) / g)^2
Simplifying:
Y = (Vo^2 * sin^2(theta)) / g - (Vo^2 * sin^2(theta)) / (2g)
Y = Vo^2 * (sin^2(theta) - sin^2(theta) / 2) / g
Y = (Vo^2 * sin^2(theta)) / (2g)
Substituting the given values:
Y = (20.9^2 * sin^2(39)) / (2 * 9.81)
Y ≈ 15.8 m
Therefore, the ball's highest point during its flight is approximately 15.8 meters.