Question

Please thoroughly explain each step, and how you got the answer. Thank you

A car is traveling with a speed of 40.0m/s up a hill that has a slope of 20.0 degrees above the horizontal when the driver realizes that he is near the edge of a cliff. For some reason the car will not get out of cruise control, so he reaches the edge of the cliff with the same speed. The perfectly vertical cliff is 32.0 m high and has a large lake at its base. What is the resultant speed of the car when it plunges into the lake? AND at what angle to the ground does this occur? (Air resistance is negligible for this car)

Answer 1

Using the equation of motion :

s = ut + 0.5*at^2

Solving the equation in vector form :

x i - 32 j = (40*cos(20 deg) i + 40*sin(20 deg) j)*t + 0.5*(-9.8 j)*t^2

So, x = 40*cos(20 deg)*t ------------(1)

and -32 = 40*sin(20 deg)*t - 0.5*9.8*t^2

So, t = 4.31 s

So, x = 40*cos(20 deg)*4.31 = 162 m

Now, using the equation of motion :

v = u + at

So, v = 40*(cos(20 deg) i + sin(20 deg) j) + (-9.8 j)*4.31

= 37.6 i - 28.6 j

So, resultant speed = sqrt(37.6^2 + 28.6^2)

= 47.2 m/s <--------- answer

angle made = atan(-28.6/37.6)

= -37.3 deg