A projectile is fired with an initial speed of 68 m/s at an angle of 55° above the horizontal plane. Determine the maximum height reached by the projectile.

To solve this problem, we will use the formula for the maximum height reached by a projectile, which is given by:

h = (v^2 * sin^2(θ)) / (2 * g)

where:

• v is the initial speed of the projectile,
• θ is the angle at which the projectile is launched,
• g is the acceleration due to gravity.

Given:

• v = 68 m/s,
• θ = 55°,
• g = 9.8 m/s^2.

First, we need to convert the launch angle from degrees to radians because the trigonometric functions in the formula work with radians. We can do this using the conversion factor π rad = 180°:

θ = 55° * (π rad / 180°) = 0.96 rad.

Then, we can substitute the given values into the formula:

h = (68 m/s)^2 * sin^2(0.96 rad) / (2 * 9.8 m/s^2) = 4624 m^2/s^2 * 0.81 / 19.6 m/s^2 = 3750.24 m / 19.6 s^2 = 191.24 m.

So, the maximum height reached by the projectile is approximately 191.24 meters.