If the initial velocity of an object thrown upwards is 14 m/s, then the time taken for the object to reach its highest point will be_______. (a = 9.8 m/s2)
1.43 s
Understanding Projectile Motion and Time to Reach Highest Point
This problem involves analyzing the motion of an object thrown upwards, which is a classic example of projectile motion under constant acceleration due to gravity. We are given the initial velocity of the object and the value for the acceleration due to gravity, and we need to find the time it takes for the object to reach its highest point.
Identifying Key Information and Physics Principles
When an object is thrown upwards, its velocity decreases as it moves against the force of gravity. At its highest point, the object momentarily stops before starting to fall back down. This means the final velocity of the object at the highest point is zero.
- Initial velocity (\(v_0\)) = 14 m/s (upwards)
- Acceleration due to gravity (\(a\)) = 9.8 m/s2 (downwards). Since upward direction is usually taken as positive, the acceleration acting downwards is negative: \(a = -9.8\) m/s2.
- Final velocity (\(v\)) at the highest point = 0 m/s.
- We need to find the time (\(t\)) taken to reach the highest point.
Applying the Kinematic Equation
We can use a standard kinematic equation that relates initial velocity, final velocity, acceleration, and time:
\(v = v_0 + at\)
Where:
- \(v\) is the final velocity
- \(v_0\) is the initial velocity
- \(a\) is the acceleration
- \(t\) is the time
Calculating the Time to Reach the Highest Point
Substitute the known values into the equation:
\(0 = 14 \, \text{m/s} + (-9.8 \, \text{m/s}^2) \times t\)
Now, we need to solve for \(t\):
\(0 = 14 - 9.8t\)
Add \(9.8t\) to both sides of the equation:
\(9.8t = 14\)
Divide both sides by 9.8 to find \(t\):
\(t = \frac{14}{9.8}\)
Calculating the value:
\(t \approx 1.42857 \, \text{s}\)
Rounding this to two decimal places, we get approximately 1.43 s.
Matching with the Options
Comparing our calculated time \(t \approx 1.43 \, \text{s}\) with the given options:
- 1.43 s
- 1 s
- 1.34 s
- 1.5 s
Our calculated value closely matches the first option, 1.43 s.
| Parameter | Value |
|---|---|
| Initial Velocity (\(v_0\)) | 14 m/s |
| Acceleration (\(a\)) | -9.8 m/s2 |
| Final Velocity at highest point (\(v\)) | 0 m/s |
| Time to reach highest point (\(t\)) | ? |
Using the equation \(v = v_0 + at\), we found the time taken for the object thrown upwards to reach its highest point.
Revision Table: Key Concepts in Vertical Motion
| Concept | Description | Relevant Equation Examples |
|---|---|---|
| Initial Velocity (\(v_0\)) | The velocity of the object at the start of the motion. | \(v = v_0 + at\) |
| Final Velocity (\(v\)) | The velocity of the object at the end point or a specific time. | \(v^2 = v_0^2 + 2a\Delta y\) |
| Acceleration (\(a\)) | The rate of change of velocity. For objects in free fall near Earth's surface, this is \(g \approx 9.8 \, \text{m/s}^2\) downwards. | \(\Delta y = v_0t + \frac{1}{2}at^2\) |
| Time (\(t\)) | The duration of the motion. | \(t = \frac{v - v_0}{a}\) |
| Displacement (\(\Delta y\)) | The change in position (height in vertical motion). | \(\Delta y = \frac{1}{2}(v_0 + v)t\) |
Additional Information: Understanding Projectile Motion Upwards
When an object is thrown straight upwards, it is subject to the constant downward acceleration due to gravity. This acceleration causes the object to slow down as it rises. The velocity becomes zero at the peak of the trajectory. After reaching the highest point, the object begins to fall back down, with its speed increasing in the downward direction due to the same acceleration.
- The time taken for the object to reach its highest point is equal to the time taken for it to fall back from the highest point to the initial launch height, assuming no air resistance.
- The maximum height reached depends on the initial velocity and the acceleration due to gravity. The equation for maximum height (\(\Delta y\)) can be derived from \(v^2 = v_0^2 + 2a\Delta y\) by setting \(v=0\), which gives \(\Delta y = \frac{-v_0^2}{2a}\) (note that \(a\) is negative for upward motion).
- These calculations are based on the assumption that air resistance is negligible. In reality, air resistance would affect the motion, reducing the maximum height and the time of flight.
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