A football quarterback shows off his skill by throwing a pass 45.90 m downfield and into a bucket. The quarterback consistently launches the ball at 40.00 ∘ above horizontal, and the bucket is placed at the same level from which the ball is thrown.

What initial speed is needed so that the ball lands in the bucket?

s = speed

45.9 = u T
where
u = s cos 40
so
T = 45.9 /(s cos 40) where T is time aloft

rise time is half T or t = T/2
v = Vi - g t
Vi = s sin 40
at top v = 0
s sin 40 = 9.81 t = 9.81[45.9/(2s cos 40]
or
2 s^2 sin 40 cos 40 = 9.81*45.9

remember trig now

Similarly, when you solve for m = 46, you get:

2 s^2 sin 40 cos 40 = 9.81*46
s = 21.40 m/s

You're welcome.

To find the initial speed needed so that the ball lands in the bucket, we can use the equations of projectile motion. The horizontal and vertical components of the motion can be analyzed separately.

1. Vertical Motion:
The vertical distance traveled by the ball is zero because the bucket is at the same level from which the ball is thrown. The only vertical motion is affected by gravity. We can use the equation:

y = v₀y * t + (1/2) * g * t^2

where
y = 0 (vertical distance)
v₀y = initial vertical velocity
g = acceleration due to gravity (9.8 m/s^2)
t = time of flight

Since the vertical distance is zero, we can rearrange the equation to solve for the time of flight:

0 = v₀y * t + (1/2) * g * t^2
0 = t * (v₀y + (1/2) * g * t)

Since t ≠ 0, we can solve for v₀y:

v₀y = - (1/2) * g * t

2. Horizontal Motion:
The horizontal distance traveled by the ball is 45.90 m. We can use the equation:

x = v₀x * t

where
x = horizontal distance
v₀x = initial horizontal velocity
t = time of flight

Since the initial velocity is launched at an angle of 40.00° above the horizontal, we can determine v₀x and v₀y using the trigonometric relationships:

v₀ = √(v₀x^2 + v₀y^2)
v₀x = v₀ * cos(40.00°)
v₀y = v₀ * sin(40.00°)

Now, we can substitute v₀x and v₀y into the equation for horizontal motion:

x = (v₀ * cos(40.00°)) * t

We need the time of flight, t, in terms of v₀. We can solve for t by rearranging the equation as follows:

t = x / (v₀ * cos(40.00°))

Substituting the values:

t = 45.90 m / (v₀ * cos(40.00°))

Now, we can substitute v₀y into the equation for vertical motion:

v₀y = - (1/2) * g * t

By substituting the value of t above, we get:

v₀ * sin(40.00°) = - (1/2) * g * (45.90 m / (v₀ * cos(40.00°)))

Simplifying the equation:

v₀ * sin(40.00°) = - (1/2) * g * (45.90 m) / (v₀ * cos(40.00°))

v₀ * sin(40.00°) * v₀ * cos(40.00°) = - (1/2) * g * (45.90 m)

Simplifying further and rearranging to solve for v₀:

v₀^2 * sin(40.00°) * cos(40.00°) = - (1/2) * g * (45.90 m)

v₀^2 = - (1/2) * g * (45.90 m) / (sin(40.00°) * cos(40.00°))

Finally, we can solve for v₀ by taking the square root of both sides:

v₀ = √[ - (1/2) * g * (45.90 m) / (sin(40.00°) * cos(40.00°)) ]

Calculating the value will give you the initial speed needed for the ball to land in the bucket.

To solve this problem, we can use the equations of motion for projectile motion.

First, let's analyze the horizontal motion. Since there is no acceleration in the horizontal direction (assuming no air resistance), the initial horizontal velocity will be the same as the final horizontal velocity.

The horizontal component of the initial velocity, Vx, can be found using the equation:
Vx = V * cos(theta)
where V is the initial speed and theta is the angle above horizontal.

Now, let's analyze the vertical motion. The vertical component of the initial velocity, Vy, can be found using the equation:
Vy = V * sin(theta)

Since the ball is launched from ground level and lands at the same level, the total time of flight, t, can be found by considering the vertical motion. Using the equation:
y = Vy * t + (1/2) * g * t^2
where y is the vertical distance (same as the height of the bucket), g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight.

Since the ball lands in the bucket, the horizontal distance, x, can be expressed as:
x = Vx * t

Given that x = 45.90 m, g = 9.8 m/s^2, and theta = 40.00 degrees, we can solve for V.

First, let's find the values of Vx and Vy using the given values of V and theta:
Vx = V * cos(theta)
Vy = V * sin(theta)

Now, let's solve for t using the vertical motion equation:
y = Vy * t + (1/2) * g * t^2

Since the ball lands at ground level, y = 0. Therefore, the equation becomes:
0 = Vy * t + (1/2) * g * t^2

Solving this equation will give us the value of t.

Finally, we can solve for V using the equation for x:
x = Vx * t

Plugging in the known values, we can solve for V, the initial speed needed so that the ball lands in the bucket.