A body is projected vertically up with a velocity of 58.8 m/s. After 3 s if the acceleration due to gravity of earth disappears. The distance travelled and the velocity of the body at the end of next 5 sec is

Options

A. 147 m , 29.4 m/s
B. 29.4 m, 147 m/s
C. 14.7 m , 2.94 m/s
D. 19.6 m, 240 m/s

Answer

To solve the problem step by step, we need to determine the distance traveled and the velocity of the body after a total of 8 seconds (3 seconds of upward motion and then 5 seconds of constant velocity).

### Step 1: Determine the initial conditions
The body is projected vertically upward with an initial velocity \( u = 58.8 \, \text{m/s} \).

### Step 2: Calculate the velocity after 3 seconds
Using the kinematic equation:
\[ v = u + at \]
where:
- \( u = 58.8 \, \text{m/s} \) (initial velocity),
- \( a = -g = -9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( t = 3 \, \text{s} \).

Substituting the values:
\[ v = 58.8 + (-9.8) \times 3 \]
\[ v = 58.8 - 29.4 = 29.4 \, \text{m/s} \]

### Step 3: Determine the velocity after the next 5 seconds
After 3 seconds, the acceleration due to gravity disappears, meaning the body will continue to move with a constant velocity. Therefore, the velocity remains:
\[ v = 29.4 \, \text{m/s} \]

### Step 4: Calculate the distance traveled in the next 5 seconds
Since there is no acceleration, we can use the formula for distance:
\[ y = vt \]
where:
- \( v = 29.4 \, \text{m/s} \) (constant velocity),
- \( t = 5 \, \text{s} \).

Substituting the values:
\[ y = 29.4 \times 5 = 147 \, \text{m} \]

### Final Results
- The velocity of the body after the next 5 seconds is \( 29.4 \, \text{m/s} \).
- The distance traveled during the next 5 seconds is \( 147 \, \text{m} \).