A car drive straight of the edge of a cliff that is 54m high. The police note that the point of impact is 130m from the base of the cliff. How fast was the car travelling when it went off the cliff.
the time to fall is
... t = √(2 * 54 / g)
speed off cliff = 130 m / t
To determine the speed of the car when it went off the cliff, we can make use of the principles of projectile motion.
First, let's define the known values:
- The height of the cliff, h = 54m
- The horizontal distance from the base of the cliff to the point of impact, x = 130m
Now, let's consider the equations of motion for projectiles:
1. The vertical motion equation:
h = (1/2) * g * t^2 (where t is the time of flight and g is the acceleration due to gravity, approximately 9.8 m/s^2)
2. The horizontal motion equation:
x = v * t (where v is the horizontal component of the velocity)
Since the car was driving straight off the cliff, there is no initial horizontal velocity. Therefore, v = 0.
Next, we can rearrange the horizontal motion equation to solve for t:
t = x / v
Now, substitute this value of t into the vertical motion equation:
h = (1/2) * g * (x / v)^2
Rearrange the equation to solve for v:
2 * h * v^2 = g * x^2
Finally, solve for v:
v^2 = (g * x^2) / (2 * h)
Taking the square root of both sides gives us the final velocity:
v = sqrt((g * x^2) / (2 * h))
Substituting the known values:
v = sqrt((9.8 * 130^2) / (2 * 54))
Calculating this equation gives us the speed at which the car went off the cliff.