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
Answers
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:
0.4
Explanation:
sorry I don't know the explanation. pls refer some other apps for solution