. A ball is thrown so that it just clears a 10 feet fence 60 feet away. If it left the hand 5 feet above the ground and at an angle of 60° with the horizontal, what was the initial velocity of the ball?

y = ax^2 + bx + c

y' = 2ax+b
y'(0) = tan60° = √3
so, b = √3

y = ax^2 + √3 x + c
y(0) = 5, so

y = ax^2 + √3 x + 5

y(60) = 10, so
3600a + 60√3 + 5 = 10
a = (1-12√3)/720

y = (1-12√3)/720 x^2 + √3 x + 5
= -0.027x^2 + 1.732x + 5

so,

g/(2v^2) sec^2 60° = 0.027
4.9 * 4 / 0.027 = v^2
v ≈ 27 m/s

. A ball is thrown so that it just clears a 10 feet fence 60 feet away. If it left the hand 5 feet above the ground and at an angle of 60° with the horizontal, what was the initial velocity of the ball?

To find the initial velocity of the ball, we can use the equations of motion for projectile motion.

In this case, the motion of the ball can be broken down into horizontal and vertical components.

Step 1: Calculate the time of flight:
Using the vertical component of motion, we can determine the time it takes for the ball to reach its highest point and fall back down.

The initial vertical velocity, Vyi = V₀ * sin(θ), where V₀ is the initial velocity and θ is the angle of projection.
Since the ball leaves the hand 5 feet above the ground, the initial vertical displacement, yi = 5 ft.
The final vertical displacement, yf = 0 ft.

Using the equation: yf = yi + Vyi * t - (1/2) * g * t², where g is the acceleration due to gravity, we can solve for the time of flight, t.

0 = 5 + (V₀ * sin(60°)) * t - (1/2) * (32.2 ft/s²) * t² (assuming the acceleration due to gravity is 32.2 ft/s²)

From this equation, we can solve for t.

Step 2: Calculate the horizontal displacement:
Using the horizontal component of motion, we can determine the horizontal displacement of the ball.

The initial horizontal velocity, Vxi = V₀ * cos(θ), where V₀ is the initial velocity and θ is the angle of projection.

The horizontal displacement, x, is given as 60 ft.

Using the equation: x = Vxi * t, we can solve for Vxi.

Step 3: Calculate the initial velocity:
The initial velocity, V₀, can now be calculated using the values of Vxi and Vyi.

V₀ = √(Vxi² + Vyi²)

Substituting the values of Vxi and Vyi, we can solve for V₀.

This will give us the initial velocity of the ball.

To find the initial velocity of the ball, we can use the equations of motion for a projectile.

Let's break down the question and determine the relevant information:

1. The initial vertical position (y₀) of the ball is given as 5 feet.
2. The vertical distance traveled by the ball is the height of the fence, which is 10 feet.
3. The horizontal distance traveled by the ball (x) is given as 60 feet.
4. The angle of projection (θ) with respect to the horizontal is 60°.

Now, let's find the initial velocity (v₀).

First, we need to split the initial velocity into its horizontal (v₀x) and vertical (v₀y) components:

v₀x = v₀ * cos(θ)
v₀y = v₀ * sin(θ)

The vertical component of the initial velocity will lead us to find the time the ball takes to reach its maximum height and later fall back down, as well as the time it takes to hit the ground.

Using the equation for vertical displacement:

y = y₀ + v₀y * t - (1/2) * g * t²

Where:
y₀ = Initial vertical position (5 ft)
v₀y = Initial vertical component of velocity (v₀ * sin(θ))
g = Acceleration due to gravity (32.2 ft/s²)
t = Time

Since the ball reaches its maximum height when its vertical velocity becomes zero, we have:

0 = v₀y - g * t_max

Rearranging the equation, we find:

t_max = v₀y / g

Next, we can use the time it takes for the ball to reach the ground (t_total) to find the horizontal component of the initial velocity.

t_total = 2 * t_max

Given that:
t_total = x / (v₀x)
t_max = v₀y / g

We can substitute these values into the equation:

x / (v₀x) = 2 * (v₀y / g)

Substituting v₀x and v₀y:

x / (v₀ * cos(θ)) = 2 * (v₀ * sin(θ) / g)

Now, we can rearrange the equation to isolate v₀:

v₀ = x / (2 * cos(θ) * sin(θ) / g)

Plugging in the values:

v₀ = 60 ft / (2 * cos(60°) * sin(60°) / 32.2 ft/s²)

Evaluating this expression will give us the initial velocity (v₀) of the ball.