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?
Can someone please solve?
jake
how long does it take to fall 58m?
h=1/2 g t^2
having time, now
horizontal distance=velocity*time
solve for velocity
To solve this problem, we can use the principles of motion in physics. We'll need to break down the motion of the car into two components: vertical and horizontal.
First, let's find the time it takes for the car to fall 58 meters vertically. We can use the equation for vertical motion:
y = ut + (1/2)gt^2
Where:
- y is the vertical distance traveled (58 m)
- u is the initial vertical velocity (which is 0 because the car was not moving vertically initially)
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time taken
Rearranging the equation, we can solve for time:
58 = (1/2)(-9.8)t^2
116 = -9.8t^2
t^2 = 116 / -9.8
t^2 ≈ 11.84
t ≈ √11.84
t ≈ 3.44 s
So, it takes approximately 3.44 seconds for the car to fall vertically.
Now, let's find the horizontal distance traveled by the car during this time. We know that the horizontal distance traveled is 126 meters, and the time taken is 3.44 seconds. So, we can use the equation for horizontal motion:
x = ut
Where:
- x is the horizontal distance traveled (126 m)
- u is the initial horizontal velocity (the velocity at which the car was traveling before going over the cliff)
- t is the time taken (3.44 s)
Rearranging the equation, we can solve for initial horizontal velocity:
u = x / t
u = 126 / 3.44
u ≈ 36.63 m/s
Therefore, the car was traveling at approximately 36.63 m/s when it went over the cliff.