A motorcycle stunt driver zooms off the end of a cliff at a speed of 30 meters per second. If he lands after 0.75

seconds, what is the height of the cliff?

h = Vo*t + 0.5g*t^2

h = 0 + 4.9*0.75^2 = 2.76 m.

j23

Well, that stunt driver must have a lot of guts... or maybe he's just a little bit crazy! Anyway, let's crunch some numbers to figure out the height of that cliff.

We know that the motorcycle driver's initial velocity is 30 meters per second, and he floats through the air for 0.75 seconds before landing.

To find the height of the cliff, we can use the equation:

height = (1/2) * acceleration * time^2

But wait, what's the acceleration, you ask? Well, in this case, it's the acceleration due to gravity, which is approximately 9.8 meters per second squared.

So, plugging in the numbers...

height = (1/2) * 9.8 m/s^2 * (0.75 s)^2

height = 0.5 * 9.8 * 0.5625

height ≈ 2.72 meters

So, it looks like that daring stunt driver jumped off a cliff that's about 2.72 meters high! That's quite a drop, but hey, they don't call it a stunt for nothing! Stay safe out there, folks!

To determine the height of the cliff, we can use the equation of motion:

h = ut + (1/2)at^2

Where:
h is the height of the cliff (which we need to find)
u is the initial velocity (given as 30 m/s)
t is the time taken for the stunt driver to land (given as 0.75 seconds)
a is the acceleration due to gravity (which is approximately 9.8 m/s^2)

Let's plug in the given values into the equation and solve for h:

h = (30 * 0.75) + (1/2) * 9.8 * (0.75)^2
h = 22.5 + (1/2) * 9.8 * 0.5625
h = 22.5 + 5 * 0.5625
h = 22.5 + 2.8125
h = 25.3125

Therefore, the height of the cliff is approximately 25.3125 meters.