To determine the velocity of the stone when it is caught on its way down, we can use the equations of motion for uniformly accelerated motion and consider the stone's vertical displacement.
Given: Initial velocity (u) = 30 m/s (upward) Vertical displacement (h) = 5 m Acceleration due to gravity (g) = 9.8 m/s² (downward)
When the stone is thrown vertically upwards, its final velocity (v) when it reaches its maximum height (at the highest point of the trajectory) is 0 m/s. At this point, the stone starts to fall downwards.
Using the equation of motion:
v² = u² + 2as
where: v is the final velocity u is the initial velocity a is the acceleration s is the displacement
For the stone's upward motion: u = 30 m/s v = 0 m/s a = -g (acceleration due to gravity, taking the opposite direction of the initial velocity) s = unknown
0² = (30)² + 2(-9.8)s 0 = 900 - 19.6s 19.6s = 900 s = 900 / 19.6 s ≈ 45.92 m
The stone reaches a maximum height of approximately 45.92 m.
Now, to determine the velocity when the stone is caught 5 m above the ground, we need to calculate the distance it falls from the maximum height to the catching point:
s = h = 5 m
Using the equation of motion:
v² = u² + 2as
where: v is the final velocity u is the initial velocity a is the acceleration s is the displacement
For the stone's downward motion: u = 0 m/s (starting from rest) v = unknown a = g (acceleration due to gravity) s = 5 m
v² = 0² + 2(9.8)(5) v² = 98 v = √98 v ≈ 9.9 m/s
Therefore, when the stone is caught on its way down, its velocity is approximately 9.9 m/s.