The highest wall that a projectile can clear is 17m, when the projectile is launched at an angle of 35.0 degrees above the horizontal. What is the projectile's launch speed?

at the top,

mgh=1/2 m v'^2 where v' is the initlal vertical velocity. find v' then

so vlaunch=v'/sin35

To find the launch speed of the projectile, we can use the conservation of energy principle. We know that the maximum height reached by the projectile occurs when its vertical velocity component becomes zero. At this point, all the initial kinetic energy is converted into potential energy.

The given information gives us the launch angle (35.0 degrees) and the maximum height reached by the projectile (17m). However, we still need to find the launch speed.

Using the conservation of energy, we can equate the initial kinetic energy to the potential energy at the maximum height:

1/2 mv^2 = mgh

Where:
m = mass of the projectile (which cancels out on both sides)
v = launch speed of the projectile
g = acceleration due to gravity (approximately 9.8 m/s²)
h = maximum height reached by the projectile

Rearranging the equation, we get:

v^2 = 2gh

Now, we can substitute the values:

v^2 = 2 * 9.8 m/s² * 17m

v^2 = 333.2 m²/s²

Taking the square root of both sides:

v ≈ √333.2

v ≈ 18.3 m/s

Therefore, the projectile's launch speed is approximately 18.3 m/s.