Question

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 3.82 m/s2 for a distance of 60.0 m to the edge of the cliff, which is 50.0 m above the ocean.
Required:
Find the car’s position relative to the base of the cliff when the car lands in the ocean.

Answer 1

Answer:

The position is  $P = 47.4 \ m$ relative to the base of the ocean

Explanation:

From the question we are told that

    The angle made by the incline with the horizontal is $\theta = 24.0 ^o$

    The constant acceleration is $a = 3.82 \ m/s^2$

    The distance covered is $d = 60.0 \ m$

    The height of the cliff is $h = 50 .0 \ m$

The velocity of the car is mathematically represented as

      $v^2 = u^2 + 2ad$

The initial velocity of the car is  u= 0

So

     $v^2 = 2ad$

substituting values

     $v^2 = 2 * 3.82 * 60$

    $v = 21.4 \ m/s$

The vertical component of this velocity is

    $v_v = -v * sin(\theta )$

substituting values

    $v_v = -21.4 * sin(24.0)$

    $v_v = -8.7 \ m/s$

The negative sign is because is moving in the negative direction of the y-axis

The horizontal  component of this velocity is

     $v_h = v * cos (\theta)$

    $v_h = 21.4 * cos (24.0)$

    $v_h = 19.5 \ m/s$

Now according to equation of motion we have

     $h = v_v*t - \frac{1}{2} * g t^2$

substituting  values

    $50 = -8.7 t - \frac{1}{2} * 9.8 t^2$

    $4.9t^2 +8.7t -50 = 0$

using quadratic equation we have that

  $t_1 = 2.42\ s \ and\ t_2 = -4.20\ s$

given that time cannot be negative

      $t = 2.42 \ s$

The  car’s position relative to the base of the cliff when the car lands in the ocean is mathematically evaluate as

           $P = v_h * t$

substituting values

          $P = 19.5 * 2.43$

         $P = 47.4 \ m$