A body is projected vertically up with a velocity 98m/s after 2sec if the acceleration disappears the velocity of the body at the end of 3sec??

To find the velocity of the body at the end of 3 seconds, we need to consider the time, initial velocity, and acceleration.

Given:
Initial velocity (u) = 98 m/s
Time (t) = 3 sec

Since the acceleration disappears after 2 seconds, we can assume that the velocity remains constant after that point.

Step 1: Find the time for which the body was accelerated.
The body was accelerated for 2 seconds, so the remaining time is 3 - 2 = 1 second.

Step 2: Find the displacement during the accelerated period.
The displacement during the accelerated period can be calculated using the equation:
s = ut + 1/2at^2

Where:
s = displacement
u = initial velocity
a = acceleration (which is unknown)
t = time

For the accelerated period (2 seconds):
s1 = ut1 + 1/2at1^2
where t1 = 2 sec

Step 3: Find the final velocity after 2 seconds.
The final velocity can be calculated using the equation:
v = u + at

For 2 seconds:
v1 = u + at1

Step 4: Find the velocity after 3 seconds.
Since the acceleration disappears after 2 seconds, the final velocity after 2 seconds is the same as the velocity after 3 seconds.

So, v2 = v1

Now, let's calculate:
Given:
u = 98 m/s
t1 = 2 sec

Step 2:
s1 = ut1 + 0.5at1^2
s1 = 98(2) + 0.5a(2^2)
s1 = 196 + 2a

Step 3:
v1 = u + at1
v1 = 98 + a(2)

Since the acceleration disappears, a = 0

Step 4:
v2 = v1 = 98 + 0(2)
v2 = 98 m/s

Therefore, the velocity of the body at the end of 3 seconds is 98 m/s.

To determine the velocity of the body at the end of 3 seconds, we need to analyze the motion of the body.

Given:
Initial velocity, u = 98 m/s (upwards)
Time, t = 3 seconds

We can break down the motion into two phases:
Phase 1: Accelerated Motion (t = 0 to 2 seconds)
Phase 2: Free Fall (t = 2 to 3 seconds)

Phase 1: Accelerated Motion (u = 98 m/s, t = 2 seconds)

In this phase, the body is projected vertically upwards with an initial velocity of 98 m/s for 2 seconds. To find the maximum height reached during this phase, we can use the equation of motion:

v = u + at

where:
v = final velocity
u = initial velocity
a = acceleration
t = time

Given that acceleration is disappearing after 2 seconds, we can assume that the acceleration during this phase is constant and equal to the acceleration due to gravity, g = 9.8 m/s^2 (downwards).

Let's calculate the maximum height reached during this phase:

v = u + at
v = 98 + (9.8)(2) [acceleration multiplied by 2 seconds]
v = 98 + 19.6
v = 117.6 m/s (upwards)

Phase 2: Free Fall (t = 2 to 3 seconds)

In this phase, the body is in free fall due to gravity. Since the acceleration is zero, the only force acting on the body is gravity.

To find the velocity at the end of the time interval (t = 3 seconds), we can use the equation:

v = u + at

where:
v = final velocity
u = initial velocity
a = acceleration
t = time

Given that the initial velocity for this phase is the final velocity from the previous phase (117.6 m/s upwards) and the acceleration is zero, we can compute the final velocity:

v = 117.6 + 0(3 - 2) [acceleration multiplied by (3 - 2)]
v = 117.6 m/s (upwards)

Therefore, the velocity of the body at the end of 3 seconds is 117.6 m/s upwards.

Please post question with all original punctuation.

Assuming the question is:
A body is projected vertically up with a velocity 98m/s. After 2sec, if the acceleration disappears, what is the velocity of the body at the end of 3sec??

From the question, it is interpreted that the body is projected with initial velocity of 98 m/s (on earth) and no further propulsion is applied, making it a vertical projectile.

After two seconds, the velocity is reduced to:
v1=v0-at=98-2*9.8=78.4.

If g is removed (by whatever means imaginable), then the body will continue at constant velocity v1=78.4 m/s upwards after 2 seconds until g is reinstated. (review Newton's first law).