Derivative/Calculus question

Question

Calculus Question - Please help
An object is thrown from a height H above an uneven ground described by a height function h(x). Suppose that the object can be thrown with maximum velocity v at an angle forming a slope m > 0. The aim is to find the slope m that will maximise the range of the throw, defined as the horizontal distance R(m) travelled by the object before impact.

It is assumed that the ground is of such shape that all of its surface can be impacted by the object, there is no sheltered spot. From Newton’s law, the trajectory of the object before impact is

r(t)= (x(t)/y(t)) = (vcos( θ)t/-g*t^2/2+vsin( θ)*t+H)

where θ = arctan(m) is the angle of the throw with respect to the x axis, and g is
the gravitational acceleration


(a) Given the parametric curve r(t), show that the object’s trajectory
before impact is described by the function
y(x) = -1/2 * g/v^2 (1+m^2) * x^2 + mx + H

(b) Use the condition for impact y(R) = h(R) and implicit differentiation
to show that the maximum range that can be obtained by varying m is
such that
R = v^2/g * 1/m

Answer 1

The parametric equations for this problem should be written as r(t) = x(t)i + y(t)j = <x(t), y(t)>. x(t)/y(t) is not r(t).
x(t) = vcosθt
y(t) = -½gt^2 + vsinθt + H

Here is the outline to solve this problem

1. To get answer for a, solve fot t in terms of x, v and θ.  Substitute this expression for t into y(t) and simplify, you get y(x)
2. Set y(R) = H for the answer in a and find dx/dm by differentiating the new equation with respect to m implicitly. You will need to use the product rule and the chain rule for differentiation to get the right answer.
3. Set dx/dm = 0 to get the answer for b.