Question
A motorbike is driven off a cliff at a horizontal velocity of 25 meters per second and takes 5 seconds to reach the ground below. Calculate the height of the e
Answers
GPT 3.5
dge of the cliff.
First, we need to find the vertical component of the velocity when the motorbike reaches the ground. We know that the acceleration due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction of the velocity).
Using the equation of motion:
vf = vi + at
where:
vf = final velocity (0 m/s when the motorbike reaches the ground)
vi = initial velocity (0 m/s as the vertical component of velocity starts at 0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
0 = 0 + (-9.8)t
t = 5 seconds
Now we can find the vertical distance traveled by the motorbike using the equation of motion:
d = vit + (1/2)at^2
where:
d = vertical distance traveled
vi = initial vertical velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (5 seconds)
So, d = 0 + (1/2)(-9.8)(5)^2
d = (1/2)(-9.8)(25)
d = -122.5 meters
Since the motorbike was driven off horizontally, the horizontal distance traveled at 25 m/s for 5 seconds would be:
25 m/s * 5 s = 125 meters
Therefore, the total distance traveled by the motorbike before hitting the ground would be the hypotenuse of a right triangle formed by the vertical and horizontal distances:
sqrt((125)^2 + (-122.5)^2) = sqrt(15625 + 15062.5) = sqrt(30687.5) = 175.1 meters
So, the height of the edge of the cliff would be 175.1 meters.
First, we need to find the vertical component of the velocity when the motorbike reaches the ground. We know that the acceleration due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction of the velocity).
Using the equation of motion:
vf = vi + at
where:
vf = final velocity (0 m/s when the motorbike reaches the ground)
vi = initial velocity (0 m/s as the vertical component of velocity starts at 0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
0 = 0 + (-9.8)t
t = 5 seconds
Now we can find the vertical distance traveled by the motorbike using the equation of motion:
d = vit + (1/2)at^2
where:
d = vertical distance traveled
vi = initial vertical velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (5 seconds)
So, d = 0 + (1/2)(-9.8)(5)^2
d = (1/2)(-9.8)(25)
d = -122.5 meters
Since the motorbike was driven off horizontally, the horizontal distance traveled at 25 m/s for 5 seconds would be:
25 m/s * 5 s = 125 meters
Therefore, the total distance traveled by the motorbike before hitting the ground would be the hypotenuse of a right triangle formed by the vertical and horizontal distances:
sqrt((125)^2 + (-122.5)^2) = sqrt(15625 + 15062.5) = sqrt(30687.5) = 175.1 meters
So, the height of the edge of the cliff would be 175.1 meters.