find the formula for maximum height and find a sin function

Question

the path traveled by an object (neglecting air resistance) that is projected by an initial height of h₀ feet, an initial velocity of v₀ feet per second and an initial angle theta is given by y=(-16/v₀²cos²θ)*x²+(tanθ)x + h₀

 

x is vertical distance and y is height, both in feet. i need help finding the formula for the maximum height of an object projected from ground lvl at velocity v₀ and angle θ. The total vertical distance the object travels is (1/32)*v₀2sinθ

the maximum height happens at half this distance. substitute it in for the x in the general model for the path of a projectile and leave your answer using a sin function. 

Answer 1

First, let us fix some errors in the problem statement:

 

x is HORIZONTAL distance, not vertical distance.  The maximum horizontal distance, when launched from the ground, is the range R, given by

 

R = (v₀²/g)sin(2θ) = (v₀²/32)sin(2θ) for g = 32 ft/s²

 

Note the formula for range given in the problem statement is incorrect.

 

Now evaluate y at R/2, and that will give maximum height y_max.

 

Remembering that sin(2θ) = 2sinθ cosθ,

 

R/2 = [v₀²/(2g)]sin(2θ) = [v₀²/(2g)](2sinθ cosθ) = (v₀²/g)sinθ cosθ

 

y(x) = -[g/(2v₀²cos²θ)]x² + (tanθ)x

 

y(R/2) = -[g/(2v₀²cos²θ)] [(v₀²/g)(sinθ cosθ)]² + (tanθ)(v₀²/g)sinθ cosθ

 

Simplify, using tanθ = sinθ / cosθ, and get

 

y(R/2) = v₀²sin²θ / (2g)

 

y_max = v₀²sin²θ / (2g)

 

That is the equation you want.

 

For g = 32 ft/s²,

 

y_max = (v₀²sin²θ)/64, with y_max in ft, v₀ in ft/s