A car drives straight off the edge of a cliff that is 58 m high. The police at the scene of the accident note that the point of impact is 126 m from the base of the cliff. How fast was the car traveling when it went over the cliff?
The height of the bldg and the point
where the car hits the ground form the ver and hor sides of a rt triangle.
The distance is = to the hyp.
tanA = y / x,
tanA = 58 / 126 = 0.4603,
A =24.7 deg,
d = x /cosA = 126 / cos24.7 =138.7 m.
V^2 = 2gd,
V^2 = 2 * 9.8 * 138.7 = 2718.5,
V = sqrt(2718.5) = 52.1 m/s.
^wrong!
To determine the speed of the car when it went off the cliff, we can use the principles of projectile motion and kinematic equations. The key information we need is the vertical displacement (the height of the cliff) and the horizontal displacement (the distance from the base of the cliff).
Let's assume the initial speed of the car when it went off the cliff is denoted as "v." To find the speed, we need to calculate the time it takes for the car to fall from the top of the cliff to the point of impact.
First, we can calculate the time taken for the car to fall vertically using the formula:
h = (1/2) * g * t^2
where
h = Vertical displacement = 58 m (height of the cliff)
g = Acceleration due to gravity = 9.8 m/s^2
t = Time taken for the car to fall
Rearranging the equation, we get:
2h = g * t^2
t^2 = (2h) / g
t = √((2 * 58) / 9.8)
Now, let's calculate the time taken for the car to travel horizontally using the formula:
d = v * t
where
d = Horizontal displacement = 126 m (distance from the base of the cliff)
v = Initial speed of the car = ? (we need to find this)
Rearranging the equation, we get:
v = d / t
Now, substituting the values of d and t, we can find the initial speed of the car:
v = 126 / (√((2 * 58) / 9.8))
Evaluating this equation will give us the speed of the car when it went off the cliff.