Q2.9 Trajectory curvature (T or H)
A ball is launched with speed v0 at an angle θ to the horizontal. The little portion of its trajectory around its highest point can be thought of as part of a circle of radius r. Find an expression for r.Hint
Reveal

Hint
Think about the acceleration at right angles to the trajectory, and make use of Key Point 1.8.Solution
Reveal

Solution
Any small enough portion of the trajectory can be thought of as a part of a circle. The effective radius of that circle (the radius of curvature of the trajectory in that little interval) is related to the acceleration at right angles to the trajectory (the centripetal acceleration) ar and the speed v by
![\[ a_r = \frac{v^2}{r} \]](mastermathpng-0.png)
Now we know that the total acceleration, , has magnitude g, and
points downwards. What makes the portion around the highest point
special is that we know that for this portion this
‘downward’ direction is at right angles to the path. Thus (at this point) we may set
![\[ g= \frac{[v_0\cos{\theta}]^2}{r} \]](mastermathpng-2.png)
![\[ r= \frac{[v_0\cos{\theta}]^2}{g} \]](mastermathpng-3.png)
You can do this question without appealing to the result for centripetal acceleration: then you have to think about the equation for the trajectory (Equation 1.10) and show that, near its highest point, it reduces to the equation of a circle with this radius. That sounds harder …and it is!