A stone is projected vertically up from the ground with velocity `40 ms^(-1)` . The interval of time between the two instants at which the stone is at a height of 60 m above the ground is `(g = 10 "ms"^(-2))`

Options

A. 4 s
B. 6 s
C. 8 s
D. 12 s

Answer

To solve the problem, we will use the equation of motion for an object projected vertically upwards. The equation we will use is:

\[ h = ut - \frac{1}{2}gt^2 \]

where:
- \( h \) is the height (60 m in this case),
- \( u \) is the initial velocity (40 m/s),
- \( g \) is the acceleration due to gravity (10 m/s²),
- \( t \) is the time in seconds.

### Step-by-Step Solution:

**Step 1: Write down the equation of motion.**

Using the equation of motion, we have:

\[ h = ut - \frac{1}{2}gt^2 \]

Substituting the known values:

\[ 60 = 40t - \frac{1}{2}(10)t^2 \]

**Step 2: Simplify the equation.**

This simplifies to:

\[ 60 = 40t - 5t^2 \]

Rearranging gives us:

\[ 5t^2 - 40t + 60 = 0 \]

**Step 3: Divide the entire equation by 5.**

To simplify further, divide the entire equation by 5:

\[ t^2 - 8t + 12 = 0 \]

**Step 4: Factor the quadratic equation.**

Now we will factor the quadratic equation:

\[ (t - 6)(t - 2) = 0 \]

**Step 5: Solve for \( t \).**

Setting each factor to zero gives us:

1. \( t - 6 = 0 \) → \( t = 6 \) seconds
2. \( t - 2 = 0 \) → \( t = 2 \) seconds

**Step 6: Find the time interval.**

The interval of time between the two instants at which the stone is at a height of 60 m is:

\[ \Delta t = t_1 - t_2 = 6 - 2 = 4 \text{ seconds} \]

### Final Answer:

The interval of time between the two instants at which the stone is at a height of 60 m above the ground is **4 seconds**.

---