A car accidently roll of a cliff. As it leaves the cliff it has horzontal velocity of 13 m/s it hits the ground 60m from the shoreline. Calculate the hight of the cliff

The height of the cliff is 104.59 m

Explanation:

The horizontal speed of the car when it leaves the cliff is 13 m/s, and it hits the ground 60m from the shoreline.

Here we can use the relationship:

Time*Speed = Distance.

To find the time that the car is in the air, we know that:

speed = 13m/s

distance = 60m

time = T

13m/s*T = 60m

T = (60m)/13m/s = 4.62 s

This means that the car is falling for 4.62 seconds.

Now let's analyze the vertical problem.

As the car leaves the cliff, it only has horizontal velocity, this means that the vertical initial velocity will be zero

The only force acting in the vertical axis is the gravitational force, this means that the acceleration will be equal to the gravitational acceleration, which is:

g = 9.8m/s^2

then:

a = -9.8m/s^2

Where the negative sign is because the acceleration is pulling the car downwards.

To get the vertical velocity, we could integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the constant of integration and the initial vertical velocity, that we already know that is equal to zero, then the vertical velocity as a function of time can be written as:

v(t) = (-9.8m/s^2)*t

To get the vertical position equation, we need to integrate again over the time:

P(t) = (1/2)*(-9.8m/s^2)*t^2 + H

Where H is the constant of integration and the initial vertical position, then H will be the height of the cliff.

We know that the car needs 4.62 seconds to hit the ground, this means that:

P(4.6s) = 0m

Then:

P(t) = (1/2)*(-9.8m/s^2)*(4.62s)^2 + H = 0

            (-4.9m/s^2)*(4.62s)^2 + H = 0

          H =  (4.9m/s^2)*(4.62s)^2 = 104.59 m

This means that the cliff is 104.59 meters high

true is the correct answer

hope i