A ball is thrown from the top of a hill at $32$ m/s, $60$ degrees above the horizontal. The hill slopes downwards at a constant angle of $30$ degrees below the horizontal. The acceleration vector is modeled by $\vec a(t) = (-0.2)\vec i + (0.2t - 10)\vec j.$ What is the maximum distance the ball is above the hill?
- Initially I thought you could find the turning point of the ball and then figure out how high above the hill it is. However this doesn't work because the instant after the turning point, the ball is coming down slower than the hill is going down, and such the ball hill is still decreasing at a quicker rate than the ball, so that wouldn't be the maximum height.
- Then I checked the answer key which shows the solution of integrating the vertical velocity to find a vertical displacement function, then adding $\tan(30) * t$. But I don't think it makes sense. One if you wanted to get the vertical displacement of the hill would you not add to $\sin(30)*t$. The second reason this doesn't make sense is because t does not progress a the same horizontal rate for the hill and the ball. For example, at $t=2$, $\cos(-30) \neq$ the integration of horizontal velocity. I'm not completely sure but I assume that the marking key is just dead wrong, however I'd like this to be confirmed by someone smarter than me. However, I also am entertaining the idea that the solution does make sense and I'm just not conceptualizing it right now.
- After some thinking I went to my original solution, and concluded that if you just find the time at which the vertical velocity starts to decrease at the same rate as the hill, then you know before that point that the hill was decreasing quicker than the ball was, and such you know that that must be the maximum distance between the hill and the ball. You could do this by finding the vertical velocity formula for the ball and equating it to $\sin(-30)$. However I'm still not sure if this would work, but I can't really explain why.
By integrating the acceleration vector $\vec a(t)=(-0.2)\vec i+(0.2t-10)\vec j$ twice and using the initial velocity vector $$\vec v_0=32\cos60^{\circ}\vec i+32\sin60^{\circ}\vec j=16\vec i+16\sqrt 3\vec j$$ and the initial position vector $s_0=0\vec i+0\vec j$ we have the velocity vector $$\vec v(t)=(-0.2t+16)\vec i+(0.1t^2-10t+16\sqrt 3)\vec j$$ and the position vector $$\vec s(t)=(-0.1t^2+16t)\vec i+(\frac{0.1}{3}t^3-5t^2+16\sqrt 3t)\vec j.$$ Hence, $x(t)=-0.1t^2+16t$ and $y(t)=\frac{0.1}{3}t^3-5t^2+16\sqrt 3t$. As commented by WWI we are asked to maximize $h(t)=y(t)+\tan30^{\circ}x(t)$ rather than $y(t)$. We have the following expression for $h(t):$ $$h(t)=\frac{0.1}{3}t^3-(5+\frac{0.1}{\sqrt3})t^2+(16\sqrt 3+\frac{16}{\sqrt3})t.$$ It is maximized at $t\approx 3.79526$ with $h_{\max}\approx 69.2071.$