An object falls freely from rest on a planet where the acceleration due to gravity is 18 m/s². After 3.9 s, what will be its speed? Answer in units of m/s.

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Chapter1: Units, Trigonometry. And Vectors
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**Physics Problem: Free Fall on a Different Planet**

An object falls freely from rest on a planet where the acceleration due to gravity is \(18 \, \text{m/s}^2\).

After \(3.9 \, \text{s}\), what will be its speed?  
Answer in units of \(\text{m/s}\).

**Solution Explanation:**

To find the speed of the object after \(3.9 \, \text{s}\), use the formula for velocity in free fall:

\[ v = u + at \]

where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity (which is 0 since the object starts from rest),
- \( a \) is the acceleration (\(18 \, \text{m/s}^2\) in this case),
- \( t \) is the time in seconds (\(3.9 \, \text{s}\)).

Plugging in the values:

\[ v = 0 + (18 \, \text{m/s}^2)(3.9 \, \text{s}) \]
\[ v = 70.2 \, \text{m/s} \]

Thus, the speed of the object after \(3.9 \, \text{s}\) will be \(70.2 \, \text{m/s}\).
Transcribed Image Text:**Physics Problem: Free Fall on a Different Planet** An object falls freely from rest on a planet where the acceleration due to gravity is \(18 \, \text{m/s}^2\). After \(3.9 \, \text{s}\), what will be its speed? Answer in units of \(\text{m/s}\). **Solution Explanation:** To find the speed of the object after \(3.9 \, \text{s}\), use the formula for velocity in free fall: \[ v = u + at \] where: - \( v \) is the final velocity, - \( u \) is the initial velocity (which is 0 since the object starts from rest), - \( a \) is the acceleration (\(18 \, \text{m/s}^2\) in this case), - \( t \) is the time in seconds (\(3.9 \, \text{s}\)). Plugging in the values: \[ v = 0 + (18 \, \text{m/s}^2)(3.9 \, \text{s}) \] \[ v = 70.2 \, \text{m/s} \] Thus, the speed of the object after \(3.9 \, \text{s}\) will be \(70.2 \, \text{m/s}\).
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