An object is thrown vertically upwards and rises to a height of 10 m. Calculate
(i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point.

Explanation

To solve the problem step by step, we will break it down into two parts: (i) calculating the initial velocity (U) with which the object was thrown upwards, and (ii) calculating the time taken (T) by the object to reach the highest point.

### Step 1: Calculate the initial velocity (U)

We can use the third equation of motion, which states:

\[ V^2 = U^2 + 2a s \]

Where:
- \( V \) = final velocity (0 m/s at the highest point)
- \( U \) = initial velocity (what we want to find)
- \( a \) = acceleration (which will be -g, where g = 9.8 m/s²)
- \( s \) = height (10 m)

At the highest point, the final velocity \( V \) is 0, so we can rewrite the equation as:

\[ 0 = U^2 + 2(-g)(s) \]

Substituting the values:

\[ 0 = U^2 - 2 \cdot 9.8 \cdot 10 \]

This simplifies to:

\[ U^2 = 2 \cdot 9.8 \cdot 10 \]

Calculating the right side:

\[ U^2 = 196 \]

Now, taking the square root to find U:

\[ U = \sqrt{196} = 14 \, \text{m/s} \]

### Step 2: Calculate the time taken (T)

We can use the first equation of motion, which states:

\[ V = U + at \]

At the highest point, \( V = 0 \), so we can rewrite the equation as:

\[ 0 = U - gt \]

Rearranging gives us:

\[ gt = U \]

Thus, we can solve for time \( t \):

\[ t = \frac{U}{g} \]

Substituting the known values:

\[ t = \frac{14}{9.8} \]

Calculating this gives:

\[ t \approx 1.43 \, \text{seconds} \]

### Final Answers:
(i) The velocity with which the object was thrown upwards is **14 m/s**.
(ii) The time taken by the object to reach the highest point is approximately **1.43 seconds**.

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