Question

a particle is projected vertically up with speed of 10ms-1. the time after which it will pass through the point 3.2m above the point of projection

Answer 1

Answer:          t = 0.4 s   is the required time

The particle is projected Vertically upward.

Initial Velocity = Vi = 10ms-1

Final Velocity = Vf = 0 ms-1  (at Maximum Height)

Time to Reach the Maximum Height:

Vf =Vi + at

Vf =Vi + gt

0  = 10 + (-10)t          ( ∵ g = - 10 m/s² upward )

10 + (-10)t = 0

- 10 t = -10

   t = 1 sec       is the time to reach the Maximum Height.

Maximum Height:

S = Vit + 1/2 gt²

S = (10)(1) + 1/2(-10)(1)²

S = 10 - 5

S = 5 m

Time to Reach at a Point 3.2 m above the Point of Projection:

S = Vit + 1/2 gt²

3.2 = (10)t + 1/2(-10)t²

3.2 = 10t -5t²

-5t² + 10t = 3.2  

-5t² + 10t - 3.2 = 0

5t² - 10t + 3.2 = 0

Using Quadratic Formula, we have

t = 1.6 s  (Not Possible)          and       t = 0.4 s

so

t = 0.4 s   is the required time

Answer 2

Answer:

0.4

Explanation:

sorry I don't know the explanation. pls refer some other apps for solution