Verified answer

Let the velocity of rain w.r.t ground is given in vector notation as

Vx i + Vy j

Now Velocity of rain w.r.t driver moving north

( draw N-S and W-E normally then rotate counterclockwise by 90 degree. This is how i'm solving this problem)

V rain,driver_North= ( Vx - (-25) ) i + Vy j

V rain,driver_South= (Vx - 25 ) i + Vy j

You know that Vy/ Vx+ 25 = tan 38.......... (1)

also since while driving south the angle is 90 hence

Vy/ Vx-25 = tan 90 => Vx-25 =0 or Vx= 25m/s

From (1) you may calculate Vy = tan 38 * 2 * 25

So velocity of rain w.r.t ground has been found out to be

V rain,ground= 25 i + 50 * tan38 j

therefore

angle it makes is tan theta= 50 * tan38/25 => tan(theta)= 2*tan 38

v = 105 m/s a= 37 degrees vx = v * Cos(a) = 83.8567 m/s vy = v * Sin(a) = 63.1906 m/s tup = vy / 9.8 =6.448 s maximum height: h = 285 + Vy * tup + (1/2) * -9.8 * tup ^ 2 = 488.727 m tdown = Sqr(2 * h / 9.8)= 9.987 s a) T = tup + tdown = 16.435 s b) Range = Vx * T = 1378.19 m = 1.378 km c) vx =83.8567 i vyf = -9.8 * tdown = -97.8726 j m/s d) Vf = Sqr(vx ^ 2 + vyf ^ 2) = 128.884 m/s Angle = atan(vyf/vx) = 310.59 degrees e) 360-310.59 = 49.41 below horizontal f) height above cliff = h - 285 488.727-285 = 203.727 m g) vector magnitude 1.438 m/s Angle =14.236 degrees