Answer
(a) Minimum speed, $u_{min}=\approx17.83 m/s$
(b) Professor will fall in the river at horizontal of $x=28.45m$ from left ramp.
Work Step by Step
(a) Using equation of trajectory, $y=xtan\theta-\frac{gx^2}{2u^2cos^2\theta}$
At $x=40m, y=-15m$ .
So, $-15=40tan53^0-\frac{9.8\times40^2}{2u^2cos^253^0}$
So minimum speed, $u_{min}= \sqrt\frac{9.8\times40^2}{2cos^253^0(40tan53^0+15)}\approx17.83 m/s$
(b) For speed half of minimum required i.e. $u=\frac{u_{min}}{2}\approx8.92 m/s$ and $y=-100m$, the value of x was found to be, $x=28.45m$.
So professor will fall in the river at horizontal of $x=28.45m$ from left ramp.
