An object is thrown vertically upward with a speed of 30 m/s. The velocity of the object half-a-second before it reaches the maximum height is

Options

A. 4.9m/s
B. 9.8m/s
C. 19.6m/s
D. 25.1m/s

Answer

To solve the problem step-by-step, we will follow the principles of kinematics, specifically focusing on the motion of an object thrown vertically upward.

### Step 1: Understand the Problem
An object is thrown vertically upward with an initial velocity (U) of 30 m/s. We need to find the velocity of the object half a second before it reaches its maximum height.

### Step 2: Identify Key Variables
- Initial velocity (U) = 30 m/s (upward)
- Acceleration (A) = -9.8 m/s² (downward, due to gravity)
- Time to reach maximum height (T) = ?

### Step 3: Calculate Time to Reach Maximum Height
At the maximum height, the final velocity (V) is 0 m/s. We can use the equation of motion:
\[ V = U + AT \]
Substituting the known values:
\[ 0 = 30 + (-9.8)T \]
Rearranging gives:
\[ 9.8T = 30 \]
\[ T = \frac{30}{9.8} \approx 3.06 \text{ seconds} \]

### Step 4: Determine Time Half a Second Before Maximum Height
The time half a second before reaching maximum height is:
\[ T_1 = T - 0.5 = 3.06 - 0.5 = 2.56 \text{ seconds} \]

### Step 5: Calculate Velocity at Time T1
Now we need to find the velocity at \( T_1 = 2.56 \) seconds using the same equation of motion:
\[ V = U + AT_1 \]
Substituting the values:
\[ V = 30 + (-9.8)(2.56) \]
Calculating:
\[ V = 30 - 25.088 \]
\[ V \approx 4.912 \text{ m/s} \]

### Conclusion
The velocity of the object half a second before it reaches the maximum height is approximately **4.9 m/s**.

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