A quarterback throws the football to a stationary receiver who is 15.2 m down the field.
The football is thrown at an initial angle of
42
◦
to the ground.
The acceleration of gravity is 9.81 m/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
See previous post.
whta
To find the initial speed at which the quarterback must throw the ball, we can use the equations of projectile motion.
1. Break down the initial velocity of the ball into its horizontal and vertical components. The horizontal component (Vx) remains constant throughout the motion, while the vertical component (Vy) changes due to gravity.
Vx = Vi * cos(theta)
Vy = Vi * sin(theta)
Where Vi is the initial velocity and theta is the angle of projection (42 degrees).
2. Use the equation of motion for the vertical component to find the time of flight (t) of the ball. We assume the ball is caught at the same height it was thrown, so the vertical displacement (y) is 0.
y = Vy * t - (1/2) * g * t^2
Since y is 0, the equation becomes:
0 = Vi * sin(theta) * t - (1/2) * g * t^2
3. Rearrange the equation to solve for t:
(1/2) * g * t^2 = Vi * sin(theta) * t
(1/2) * g * t = Vi * sin(theta)
t = (2 * Vi * sin(theta)) / g
4. Use the equation for the horizontal component of the motion to find the horizontal distance (x) traveled by the ball:
x = Vx * t
5. The receiver is located 15.2 meters down the field, which is equal to the horizontal distance traveled by the ball (x). Therefore,
x = 15.2 m
6. Substitute the known values into the equations:
15.2 m = Vi * cos(theta) * [(2 * Vi * sin(theta)) / g]
7. Solve the equation for Vi:
Vi * cos(theta) * [(2 * Vi * sin(theta)) / g] = 15.2 m
Vi^2 * cos(theta) * sin(theta) = (15.2 m * g) / 2
Vi^2 = (15.2 m * g) / [2 * cos(theta) * sin(theta)]
Vi = sqrt[(15.2 m * g) / [2 * cos(theta) * sin(theta)]]
8. Substitute the given values:
θ = 42 degrees
g = 9.81 m/s^2
Vi = sqrt[(15.2 m * (9.81 m/s^2)) / [2 * cos(42 degrees) * sin(42 degrees)]]
9. Calculate:
Vi = sqrt[(149.11272) / (0.755113) ]
Vi = sqrt[197.60]
Vi = 14.04 m/s
Therefore, the initial speed at which the quarterback must throw the ball for it to reach the receiver is 14.04 m/s.
To find the initial speed at which the quarterback must throw the ball, we can use the kinematic equations of motion for projectile motion.
First, let's analyze the motion of the ball horizontally and vertically separately.
Horizontal Motion:
The football is thrown horizontally, which means there is no acceleration in the horizontal direction. Therefore, the initial horizontal velocity (Vx) remains constant throughout the motion.
Vertical Motion:
The football is thrown at an angle of 42 degrees with respect to the ground, so we need to split the initial velocity (Vi) into horizontal (Vx) and vertical (Vy) components.
Given that the acceleration due to gravity (g) is 9.81 m/s^2 and the vertical distance (dY) is 15.2 m, we can use the following equation:
dY = Vi * sin(θ) * t - (1/2) * g * t^2
where t is the time of flight (the time it takes for the ball to reach the receiver). At the maximum height, the vertical velocity (Vy) will be 0; hence, we can write:
0 = Vi * sin(θ) - g * t
We can rearrange this equation to solve for t:
t = Vi * sin(θ) / g
Now, let's consider the horizontal motion. Since there is no horizontal acceleration, the horizontal distance (dX) traveled by the ball is given by:
dX = Vx * t = Vi * cos(θ) * t
To solve for Vi, we can use the fact that the total horizontal distance traveled (dX) is equal to the given distance of 15.2 m:
dX = 15.2 m
Vi * cos(θ) * t = 15.2 m
Substituting the expression for t, we get:
Vi * cos(θ) * (Vi * sin(θ) / g) = 15.2 m
From this equation, we can solve for Vi:
Vi^2 * cos(θ) * sin(θ) / g = 15.2 m
Simplifying further:
Vi^2 * (1/2) * sin(2θ) / g = 15.2 m
Vi^2 = (15.2 m * g) / (0.5 * sin(2θ))
Now, substitute the given values of θ (42°) and g (9.81 m/s^2) into the equation and solve for Vi:
Vi^2 = (15.2 m * 9.81 m/s^2) / (0.5 * sin(2 * 42°))
Vi^2 = 150.9528 m^2/s^2
Taking the square root of both sides:
Vi = √(150.9528 m^2/s^2) = 12.29 m/s
Therefore, the initial speed at which the quarterback must throw the ball for it to reach the receiver is approximately 12.29 m/s.