math A particle A is projected vertically upwards from a point O on horizontal ground with speed 20ms. At the same instant a particle B is released from rest at point P which is 30 m above the ground. The point P Is not directly above O ignore air resistance. Find thr time that ...
To solve this problem, we first need to determine the time it takes for particle A to reach the same height as particle B.
Let's break down the motion of each particle separately:
Particle A:
- Initial speed (u) = 20 m/s (upwards)
- Acceleration (a) = -9.8 m/s^2 (due to gravity, acting downwards)
- Initial height (h) = 0 m (on the ground)
Using the kinematic equation: h = ut + (1/2)at^2, we solve for the time (t) when the height (h) is equal to the initial height of particle B:
0 = (20 * t) + (0.5 * -9.8 * t^2)
Simplifying the equation, we get:
-4.9t^2 + 20t = 0
Factorizing, we obtain:
t(-4.9t + 20) = 0
This equation gives us two solutions: t = 0 and t = 20/4.9.
However, since we are interested in the time taken for particle A to reach point P, we discard t = 0 as it represents the starting time. Therefore, the valid solution is:
t = 20/4.9 ≈ 4.08 seconds
So, it takes approximately 4.08 seconds for particle A to reach the same height as particle B.