A particle is projected from ground in a vetical direction at t=0 sec. at t=0.8 s, it reaches h=14m .It will again come to the same height at ?

h= v₀•t-g•t²/2 ,

v₀=h/t +gt/2 = 14/0.8 +9.8•0.8/2 = 21.42 m/s.
Time of the motion to the highest point
t₀=v₀/g =21.42/9.8=2.19 s.
t1=2•2.19 – 0.8=3.58 s.

To find the time at which the particle will again come to the same height, we can use the equations of motion.

The equation that relates the height (h), initial velocity (u), time (t), and acceleration due to gravity (g) is:

h = ut + (1/2)gt^2

Given:
Initial height (h) = 14m
Time (t) = 0.8s

We can rearrange the equation to solve for the initial velocity (u):

u = (h - (1/2)gt^2) / t

Substituting the given values:

u = (14 - (1/2) * 9.8 * (0.8)^2) / 0.8

Solving this equation, we get:

u ≈ 10.15 m/s

Now we can find the time it takes for the particle to reach the same height again. At the highest point, the velocity will be zero. Therefore, we can use the equation:

v = u + gt

Where v is the final velocity (which is zero at the highest point). Solving for t:

0 = 10.15 + (-9.8) * t

Rearranging the equation:

9.8t = 10.15

t ≈ 1.04 seconds

Therefore, the particle will again come to the same height (14m) approximately 1.04 seconds after it was initially projected.

To determine when the particle will reach the same height again, we can use the concepts of projectile motion and apply the equations of motion.

First, let's denote the initial speed of the particle as "u" and the time it takes to reach the same height again as "t." We know that at t = 0, the particle is projected from the ground vertically, so its initial velocity (u) is in the upward direction.

Using the equation of motion for vertical displacement:

h = ut + (1/2)gt^2

Where:
- h is the displacement in the vertical direction (14m in this case).
- u is the initial vertical velocity (which is positive since it's projected upward).
- g is the acceleration due to gravity (-9.8 m/s^2, taking downward as the positive direction).
- t is the time taken.

Substituting the given values:

14 = u(0.8) + (1/2)(-9.8)(0.8)^2

Simplifying:

14 = 0.8u - 3.2

Rearranging the equation:

0.8u = 14 + 3.2
0.8u = 17.2
u = 17.2 / 0.8
u = 21.5 m/s

So, the initial vertical velocity of the particle is 21.5 m/s.

Now, we can find the time it takes for the particle to reach the same height again using the equation:

h = ut + (1/2)gt^2

Rearranging this equation:

0 = (1/2)gt^2 + ut - h

Substituting the known values:

0 = (1/2)(-9.8)t^2 + (21.5)t - 14

This equation is a quadratic equation in terms of "t." We can solve it using the quadratic formula:

t = [-b ± √(b^2 - 4ac)] / (2a)

Here, a = (1/2)(-9.8), b = 21.5, and c = -14.

Substituting these values into the quadratic formula:

t = [-(21.5) ± √((21.5)^2 - 4(0.5)(-14))] / [2(0.5)(-9.8)]

Calculating this equation will give us two possible values of "t." The positive value will correspond to the time it takes for the particle to reach the same height again.

Once you plug in the values and solve the quadratic equation, you will find the value of "t" when the particle reaches the same height again.

Can we use 10 meter per second square instead of 9.8 meter per sec square