Question

A stunt car is being used in a movie scene. Big Arm Arnold is to drive off a 74.7 m high cliff with an initial horizontal speed if 55 km/h. To film this stunt, the director needs to know how long the car will be in freefall, and how far from the base of the cliff the stunt car will land. Also determine the stunt car’s final velocity.

Answer 1

Answer:

1) The time duration of the car in free fall is approximately 3.9 seconds

2) The distance from the base of the cliff the stunt car will land is approximately 59.6 m

3) The speed of the stunt car when it lands is approximately 41.22 m/s

Explanation:

1) The given information are;

The height of the cliff from which the big Arnold is to drive, h = 74.7 m

The initial horizontal speed of the stunt car, vₓ = 55 km/h ≈ 15.28 m/s

The time duration of the car in free fall is given by the equation for free fall, from height, h, as follows;

h = 1/2 × g × t²

Where;

g = The acceleration due to gravity = 9.81 m/s²

h = 74.7 m

Substituting the values, gives;

74.7 = 1/2 × 9.81 × t²

t² = 74.7/(1/2 × 9.81) ≈ 15.23 s²

t = √15.23 ≈ 3.9

The time duration of the car in free fall, t ≈ 3.9 s

2) The distance from the base of the cliff the stunt car will land = The horizontal speed × The time in free fall

The distance from the base of the cliff the stunt car will land = 15.28 × 3.9 ≈ 59.6 m

The distance from the base of the cliff the stunt car will land ≈ 59.6 m

3) The vertical velocity, $v_y$, due to gravity of the stunt car when it lands is given by the following free fall equation of motion;

$v_y$² = 2 × g × h

∴ $v_y$² = 2 × 9.81 × 74.7 = 1465.614

$v_y$ = √(1465.614) ≈ 38.28 m/s

The resultant speed , v, of the stunt car when it lands is given by the following equation

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{15.28^2 + 38.28^2} \approx 41.22 \ m/s$

The speed of the stunt car when it lands ≈ 41.22 m/s.