A particle is projected up with an initial velocity of `80 ft//sec`. The ball will be at a height of 96ft from the ground after

Options

A. 2.0 and 3.0 sec
B. Only at 3.0 sec
C. Only at 2.0 sec
D. After 1 and 2 sec

Answer

To solve the problem of determining when a particle projected upwards with an initial velocity of 80 ft/sec reaches a height of 96 ft, we can use the kinematic equation for uniformly accelerated motion. The equation we will use is:

\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \( s \) is the displacement (height in this case),
- \( u \) is the initial velocity,
- \( a \) is the acceleration (which will be negative due to gravity),
- \( t \) is the time.

### Step 1: Identify the variables
- Initial velocity \( u = 80 \) ft/sec
- Height \( s = 96 \) ft
- Acceleration due to gravity \( a = -32 \) ft/sec² (since it acts downwards)

### Step 2: Set up the equation
Substituting the known values into the equation:

\[ 96 = 80t - 16t^2 \]

Here, we have used \( \frac{1}{2} a = -16 \) (since \( a = -32 \) ft/sec²).

### Step 3: Rearrange the equation
Rearranging gives us:

\[ 16t^2 - 80t + 96 = 0 \]

### Step 4: Solve the quadratic equation
Now, we will use the quadratic formula to solve for \( t \):

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Where:
- \( a = 16 \)
- \( b = -80 \)
- \( c = 96 \)

Calculating the discriminant:

\[ b^2 - 4ac = (-80)^2 - 4 \cdot 16 \cdot 96 \]
\[ = 6400 - 6144 \]
\[ = 256 \]

Now substituting into the quadratic formula:

\[ t = \frac{80 \pm \sqrt{256}}{2 \cdot 16} \]
\[ = \frac{80 \pm 16}{32} \]

This gives us two possible solutions for \( t \):

1. \( t_1 = \frac{96}{32} = 3 \) seconds
2. \( t_2 = \frac{64}{32} = 2 \) seconds

### Step 5: Conclusion
The particle reaches a height of 96 ft at two different times: \( t = 2 \) seconds (on the way up) and \( t = 3 \) seconds (on the way down).