The Swiss tennis player Roger Federer hits the ball at the point r(0) = left langle 0,0,3 right rangle. The initial velocity is r ‘(0) = left langle 100,10,13 right rangle. The tennis ball exp

From the given information, we know that the initial position of the tennis ball is {eq}left langle 0,0,3 right rangle {/eq} and the initial velocity is {eq}left langle 100,10,13 right rangle {/eq}. We can find the equation of motion of the tennis ball using the kinematic equation {eq}r(t) = r(0) + r'(0)t + \frac{1}{2}at^2 {/eq} where {eq}a = left langle 2,0,-32 right rangle {/eq} is the constant acceleration experienced by the ball. The equation of motion of the ball is then {eq}r(t) = leftlangle 100t,10t,3+13t-16t^2 rightrangle {/eq} To find where the ball hits the ground, we need to find the time at which {eq}z = 3+13t-16t^2 = 0 {/eq}. This is a \quadratic equation which can be solved using the \quadratic formula, giving {eq}t = \frac{13 + sqrt{569}}{16} \approx 2.65 {/eq} (we discard the negative solution as it does not make sense in this context). To find the position of the ball at this time, we plug in {eq}t = 2.65 {/eq} into the equation of motion and get {eq}r(2.65) \approx leftlangle 265, 26.5, 0 rightrangle {/eq}, {/eq} which is where the ball hits the ground. To find the speed of the ball at that time, we need to find the magnitude of the velocity vector at {eq}t = 2.65{/eq}. Using the formula {eq}left|{mathbf{v}}right| = sqrt{{mathbf{v}}cdot{mathbf{v}}} {/eq}, we get {eq}left|{mathbf{v}}(2.65)right| \approx 41.38 {/eq}. Therefore, the speed of the ball at the time it hits the ground is about 41.38 units per second, where the exact unit is not given.