A stone is thrown vertically upward with a speed of 19.0m/s
How fast is it moving when it reaches a height of 10.0m
Ke at bottom = (1/2) m v^2 = 180.5 m
Pe at top = m g h = m (9.81)(10) = 98.1 m
Ke lost in rise = 98.1 m
so
Ke at top = (1/2)m v^2 = (180.5-98.1) m
so
(1/2) v^2 = 82.4
so
v = 12.8 m/s
To find the speed of the stone when it reaches a height of 10.0m, we can use the laws of motion and the principle of conservation of energy.
1. Use the equation of motion:
v^2 = u^2 + 2as,
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
In this case, the stone is thrown vertically upward, so the final velocity at its maximum height will be zero (v = 0 m/s), the initial velocity (u) is 19.0 m/s, the acceleration due to gravity (a) is -9.8 m/s^2 (negative because it acts downward), and the displacement (s) is 10.0 m.
Plugging in the values into the equation:
0 = (19.0 m/s)^2 + 2(-9.8 m/s^2) * 10.0 m
2. Solve for the final velocity (v):
0 = 361 m^2/s^2 - 196 m^2/s^s
196 m^2/s^2 = 361 m^2/s^2
v^2 = 165 m^2/s^2
v = sqrt(165 m^2/s^2)
v ≈ 12.85 m/s
Therefore, the stone will have a speed of approximately 12.85 m/s when it reaches a height of 10.0m.