Question

A particle is projected vertically upwards with a velocity of 34.3 m/s. For how long is the particle at a height of 49 m and above the point of projection?

Answer 1

The particle will be at a height of 49 m and above the point of projection for approximately 3.16 seconds.

Step-by-step explanation:

To calculate the time for the particle to be at a height of 49 m and above the point of projection, we can use the equations of motion. Since the particle is projected vertically upwards, we can use the equation:

h = u2 - 1/2gt2

Where h is the height, u is the initial velocity, g is the acceleration due to gravity (approx. 9.8 m/s2), and t is the time. Rearranging the equation to solve for t, we get:

t = √(2h/g)

Substituting the given values, we have:

t = √(2 * 49 / 9.8) = √10 = 3.16 seconds

Therefore, the particle is at a height of 49 m and above the point of projection for approximately 3.16 seconds.

Answer 2

Calculate the time to reach a height of 49 m:

S = V0 t - 1/2 g t^2

S = 34.3 t - 1/2 * 9.8 * t^2

4.9 t^2 - 34.3 t + 49 = 0

I get t = 2 and t = 5 as solutions of the quadratic.

We know that it takes v / a = 34.3 / 9.8 = 3.5 sec to reach maximum height.

So we have:

2 sec to reach 49 m

3.5 sec to reach max height

1.5 sec from 49 m to max height

1.5 sec from max height back down to 49 m

2 sec to fall from 49 m to ground

2 sec + 1.5 sec + 1.5 sec = 5 sec to go from ground to max height and back down to 49 m