The highest barrier that a projectile can clear is 12.7 m, when the projectile is launched at an angle of 17.5° above the horizontal. What is the projectile's launch speed?

To determine the projectile's launch speed, we can use the equations of motion for projectile motion.

First, let's analyze the motion along the vertical axis. The vertical motion can be described using the equation:
y = yo + voy * t - (1/2) * g * t^2
where:
y = vertical displacement (in this case, 12.7 m)
yo = initial vertical position (0 m)
voy = initial vertical velocity (unknown)
t = time of flight (unknown)
g = acceleration due to gravity (approximately 9.8 m/s^2)

Now, let's analyze the motion along the horizontal axis. The horizontal motion is described by the equation:
x = xo + vox * t
where:
x = horizontal displacement (unknown)
xo = initial horizontal position (0 m)
vox = initial horizontal velocity (unknown)
t = time of flight (unknown)

Since there is no acceleration along the horizontal axis, the initial horizontal velocity (vox) remains constant throughout the projectile's motion. Therefore, we can use the equation:
vox = vo * cos(theta)
where:
vo = launch speed (unknown)
theta = launch angle (17.5°) in radians (radians = degrees * pi/180)

Now, we can substitute the values into the equations. Let's start by finding the time of flight (t) using the vertical displacement equation:
y = yo + voy * t - (1/2) * g * t^2
12.7 = 0 + voy * t - (1/2) * 9.8 * t^2
0.5 * 9.8 * t^2 - voy * t + 12.7 = 0

This is a quadratic equation in terms of t. We can solve it using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
where a = 0.5 * 9.8, b = -voy, and c = 12.7.

Now, we need to find the horizontal displacement (x), which can be obtained by substituting the values into the horizontal displacement equation:
x = xo + vox * t
x = 0 + vo * cos(theta) * t

Finally, we can solve for the launch speed (vo). Using the vertical and horizontal displacements obtained from the equations above, we have:
x = vo * cos(theta) * t
vo = x / (cos(theta) * t)

Substituting the values of x, theta, and t will give us the launch speed in m/s.

To find the projectile's launch speed, we need to break down the given information and use the kinematic equations.

Given:
- Maximum height (h) = 12.7 m
- Launch angle (θ) = 17.5°

We can use the equation for the maximum height reached by a projectile:

h = (v₀² * sin²θ) / (2 * g)

where:
- v₀ is the initial velocity (launch speed)
- θ is the launch angle
- g is the acceleration due to gravity (9.8 m/s²)

Rearranging the equation, we get:

v₀ = sqrt((h * 2 * g) / sin²θ)

Now let's substitute the given values into the equation:

v₀ = sqrt((12.7 * 2 * 9.8) / sin²17.5°)

Calculating this expression yields:

v₀ ≈ 29.90 m/s

Therefore, the projectile's launch speed is approximately 29.90 m/s.

The vertical initial speed is Vsin17.5

the vertical speed at max height is zero.

Vf^2=(Vsin17.5)^2-2*9.8*12.7
solve for V, knowing Vf is zero.