What is the distance from the base of the cliff where the stunt car lands after driving off a 115-meter tall cliff at 25 m/s horizontally?
Final Answer: The car lands approximately 135 meters from the base of the cliff.
To determine the horizontal distance the stunt car travels before landing, we can utilize the kinematic equations of motion. It's important to note that while the vertical motion of the car is affected by gravity, the horizontal motion remains unaffected.
Kinematic Equations of Motion Used:
Given:
Initial vertical displacement (h) = 115 m (height of the cliff)
Initial vertical velocity (v_y) = 0 m/s (car starts from rest vertically)
Vertical acceleration (a_y) = 9.81 m/s² (acceleration due to gravity)
Using the equation:
h = v_y * t + 0.5 * a_y * t²
Solving for time (t) when the car hits the ground:
115 m = 0 * t + 0.5 * 9.81 m/s² * t²
t² = (2 * 115 m) / 9.81 m/s²
t ≈ 5.40 s
Since the car moves horizontally at a constant speed of 25 m/s, the horizontal distance (d) the car travels can be calculated using the equation:
d = v_x * t
d = 25 m/s * 5.40 s
d ≈ 135 m
Therefore, the car lands approximately 135 meters horizontally away from the base of the cliff.