differentiation to algebraic trigonometric

Question

A projectile is launched from a platform on a ballistic trajectory described by the equation:   

h= -0.01d^2+1.3d+5

where    d = the distance of the projectile from the launch site in metres

and        h = the height of the projectile above the ground in metres

 

Using differential calculus:

Determine the projectile’s rate of climb with respect to it’s distance travelled when it was launched (at launch d = 0), and from this the angle of elevation (from horizontal) at which the projectile was launched.

Determine the distance from the launch site at which the projectile reaches its maximum height.

Using the distance from the launch site at which the projectile reaches its maximum height, calculate the value of the maximum height reached by the projectile.

 

Answer 1

∂h/∂d=-0.02d+1.3 is the rate of climb as measured in terms of the horizontal distance travelled. This is the gradient=tanø, where ø is the angle of elevation. So at d=0 the rate of climb is 1.3 and tanø=1.3 initially, making the angle of elevation ø=52.43º.

The maximum height is when ∂h/∂d=0=-0.02d+1.3 so d=1.3/0.02=65m.

From this we can work out the maximum height: h=-0.01*65^2+1.3*65+5=47.25m (the vertex of the parabola).

[I used ∂ rather than d for the differential notation to avoid confusion with d for distance.]