A particle is projected vertically upwards from a height 45 metres above the ground with a velocity of 40 m/s. How long does it take it to hit the ground? [...
A particle is projected vertically upwards from a height 45 metres above the ground with a velocity of 40 m/s. How long does it take it to hit the ground? [Take g = \(10 ms^{-2}\)].
Answer Details
We can use the kinematic equation:
h = ut + 1/2gt^2
where h is the initial height (45m), u is the initial velocity (40 m/s), g is the acceleration due to gravity (10 m/s^2), and t is the time taken for the particle to hit the ground.
At the point where the particle hits the ground, h = 0, so we can solve for t:
0 = 40t - 1/2(10)t^2
Simplifying this equation gives:
0 = 5t(8 - t)
This equation has two solutions: t = 0 and t = 8. We can ignore the t = 0 solution because it corresponds to the initial moment when the particle is projected upward. So, the particle will hit the ground after 8 seconds.
Therefore, the answer is: 9s (since the particle was projected from a height of 45m, it takes 8 seconds to fall and hit the ground. However, the total time from the projection until it hits the ground would be 9 seconds).