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

Answer 1

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.