Question

A distant planet has an acceleration due to gravity of 4 m/s^2 near its surface. An object is released from rest from the top of a tall cliff on the planet, and the object lands at the bottom of the cliff in 20 seconds. A second object is then thrown upward from the edge of the same cliff with a speed of 4 m/s. How long does it take the second object to reach the bottom of the cliff?

Answer 1

We can use the formula
`\(v = u + at\)` for the second object. The object is thrown upward, so the acceleration
`(\(a\))` is negative (equal to the acceleration due to gravity).

The formula becomes:

`\[v = u - gt\]`

where:

`\(v\)` is the final velocity (0 m/s when the object reaches the bottom),

`\(u\)` is the initial velocity (4 m/s),

`\(g\)` is the acceleration due to gravity (-4 m/s²),

`\(t\)` is the time.

Now, we want to find the time
`(\(t\))` it takes for the object to reach the bottom, so we set
`\(v\)` to 0 and solve for
`\(t\)`:

`\[0 = 4 - 4t\]`

`\(0 = 4 - 4t\)` for
`\(t\)`:

`\[4 - 4t = 0\]`

Subtract 4 from both sides:

`\[-4t = -4\]`

Divide both sides by -4:

`\[t = 1\]`

So, it takes 1 second for the second object to reach the bottom of the cliff.

Answer 2

The second object takes 1 second to reach the bottom of the cliff.

Step-by-step explanation:

To calculate the time it takes for the second object to reach the bottom of the cliff, we can use the equations of motion. Since the object is thrown upward, its acceleration due to gravity is negative (-4 m/s²). We know the initial velocity (4 m/s) and we are looking for the time it takes to reach the bottom of the cliff. We can use the equation:

v = vo + at

where v is the final velocity, vo is the initial velocity, a is the acceleration, and t is the time. Rearranging the equation, we have:

t = (v - vo) / a

Plugging in the values, we get:

t = (0 - 4) / -4 = 1 second

So, it takes the second object 1 second to reach the bottom of the cliff.