A physics student stands on the edge of a cliff 2x meters. He throws his physics homework straight up in the air. Derive an expression to calculate the time taken by the homework to come back to the student.

Vf=Vi+at

Nope, time works here:

hf=hi+vi*t-1/2 *(9.8)t^2
but hf=hi
t(vi-4.9t)=0
t=vi/4.9

To derive an expression for the time taken by the homework to come back to the student, we can use the equations of motion.

First, let's define the variables:
- Initial velocity of the homework, u (upwards)
- Acceleration due to gravity, a (downwards)
- Time taken, t

Since the homework is thrown straight up, its initial velocity is positive and its acceleration, due to gravity, is negative. The acceleration due to gravity is usually denoted as "g" and is approximately equal to 9.8 m/s² on Earth.

When the homework reaches its maximum height, its final velocity will be zero. At this point, the homework will start falling downwards due to the force of gravity, and its acceleration will be -9.8 m/s².

Using the equation of motion:

v = u + at

where:
v - final velocity (0 m/s at maximum height)
u - initial velocity (upwards)
a - acceleration due to gravity
t - time taken

Substituting the values, we get:

0 = u + (-9.8)t

Simplifying the equation, we have:

9.8t = u

Now, we need to find the time it takes for the homework to reach its maximum height and then come back down to the student. At its maximum height, the final velocity is zero and the displacement is 2x meters (distance from the student to the cliff edge).

Using the equation of motion:

s = ut + (1/2)at²

where:
s - displacement (2x meters)
u - initial velocity (upwards)
a - acceleration due to gravity
t - time taken

We substitute the values:
2x = ut + (1/2)(-9.8)t²

Simplifying the equation, we have:

2x = ut - (4.9)t²

Since the homework goes up and then comes back down, the total time taken can be considered as the sum of the time taken for it to reach its maximum height (t₁) and the time taken for it to come back down (t₂).

Thus, t = t₁ + t₂

From the earlier derived equation, we have:

t₂ = t₁ = u/9.8

Substituting this in the displacement equation, we get:

2x = ut₁ - (4.9)t₁²

Substituting t₁ = u/9.8, we have:

2x = u(u/9.8) - (4.9)(u/9.8)²

Simplifying the equation, we get:

2x = (u²/9.8) - (0.25)(u²/9.8)

Multiplying through by 9.8 to eliminate the denominators, we have:

19.6x = u² - 0.25u²

Simplifying further, we get:

19.6x = 0.75u²

Solving for u², we have:

u² = (19.6x) / 0.75

Taking the square root of both sides, we get:

u = √((19.6x) / 0.75)

Finally, substituting this value for u in the expression for t₁, we have:

t₁ = u / 9.8
t₁ = (√((19.6x) / 0.75)) / 9.8

And since t = t₁ + t₂, we have:

t = (√((19.6x) / 0.75)) / 9.8 + (√((19.6x) / 0.75)) / 9.8

Simplifying the equation, we get:

t = 2(√((19.6x) / 0.75)) / 9.8

Hence, the expression to calculate the time taken by the homework to come back to the student is:

t = 2(√((19.6x) / 0.75)) / 9.8