A particle is projected vertically upwards from ground level with an initial velocity of 34.94m/s. Neglecting air resistance, calculate the total time taken, in seconds, for the particle to hit the ground again.

Take the magnitude of the acceleration due to gravity to by 9.81m/s^2.

To find the total time taken for the particle to hit the ground again, we can use the following steps:

Step 1: Determine the initial velocity and acceleration.
Given:
Initial velocity (u) = 34.94 m/s (upward)
Acceleration due to gravity (g) = 9.81 m/s^2 (downward)

Step 2: Determine the time taken to reach the highest point.
When the particle reaches the highest point, its final velocity (v) would be zero since it momentarily stops before falling back down. We can use the equation of motion:

v = u + gt

Rearranging the equation to solve for time (t):

t = (v - u) / g

Substituting the given values:

t = (0 - 34.94) / 9.81

Step 3: Calculate the time taken to reach the highest point.
Simplifying the equation:

t = -34.94 / 9.81

t ≈ -3.56 seconds

Since time cannot be negative, we ignore the negative sign and take the absolute value:

t = 3.56 seconds

Step 4: Calculate the total time taken to hit the ground.
The total time taken is twice the time taken to reach the highest point because the particle would take the same amount of time to fall back down. Therefore, the total time taken is:

Total time = 2 * t = 2 * 3.56 = 7.12 seconds

Therefore, the total time taken for the particle to hit the ground again is approximately 7.12 seconds.

Find the time to the top and double it.

v = Vi - g t
at the top, v = 0
0 = 34.9 - 9.81 t
t = 34.9/9.81
2 t = 2 * 34.9/9.81