THE HUMAN CANNONBALL - AGAIN! (2 pts)

On July 23, 1997, Washington County, in Oregon was advertising their Fair & Rodeo special attraction on the Internet: "As the countdown begins, David Smith, Jr. - THE BULLET - puts on his safety helmet, slides down the slender barrel of the cannon, positioning himself to become a human projectile. Then begins the countdown and the announcer asks the audience to cover their ears as protection from the noise of the blast. Three! Two! One! FIRE! David Smith, Jr. - THE BULLET - blasts into the air, reaching a height of 60 feet above blast off and landing safely in his pre-positioned safety net at a distance of 160 feet. This is a death-defying stunt to send shivers down your spine!"

Use the given information and assume that the tip of the cannon is at the same level as the pre-positioned safety net to find the equation of the flight of David Smith, Jr. aka THE BULLET!

To find the equation of the flight of David Smith, Jr. aka THE BULLET, we need to use the given information: the height reached by the human cannonball (60 feet) and the distance he lands from the cannon (160 feet).

Let's consider the vertical motion of the human cannonball first. We know that the initial velocity in the vertical direction is zero since the bullet starts from rest at the tip of the cannon. The only force acting on the bullet in the vertical direction is gravity. Therefore, we can use the kinematic equation to determine the height reached:

h = v0t + (1/2)gt^2

where h is the height, v0 is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.

At the maximum height, the vertical velocity is zero. So, we can determine the time it takes for the bullet to reach the maximum height using the equation:

0 = v0 + gt_max

Solving for t_max, we get:

t_max = -v0 / g

Substituting this value into the equation for height, we have:

h_max = v0 * (-v0 / g) + (1/2)g * (-v0 / g)^2

Simplifying further:

h_max = - (1/2) * (v0^2 / g)

We are given that the height reached is 60 feet, so we can substitute this value into the equation:

60 = - (1/2) * (v0^2 / g)

Next, let's consider the horizontal motion of the human cannonball. Assuming there is no air resistance, the horizontal motion is uniform. This means that the horizontal distance traveled (x) is given by:

x = v_horizontal * t

where v_horizontal is the horizontal velocity and t is the time.

Since the initial horizontal velocity is unknown, we need to find it by using the given distance of 160 feet. Considering that the time to reach the maximum height is half the total time in the air, we can express the total time as:

t_total = 2 * t_max

Substituting the expression for t_max, we get:

t_total = 2 * (-v0 / g)

The horizontal velocity can be calculated by dividing the horizontal distance by the total time:

v_horizontal = x / t_total

Substituting the given distance of 160 feet and the expression for t_total, we have:

v_horizontal = 160 / (2 * (-v0 / g))

Simplifying further:

v_horizontal = -80 / (v0 / g)

Now we have the vertical velocity v0 and the horizontal velocity v_horizontal. The equation of the flight can be expressed as:

y = h_max + v0t - (1/2)gt^2

where y is the vertical position of the bullet as a function of time t.

x = v_horizontal * t

where x is the horizontal position of the bullet as a function of time t.

These two equations describe the flight of David Smith, Jr. aka THE BULLET.