1.A stone is thrown upward with an initial velocity of 20m/s.Aire resistance is ignored.(g=10m/S^2)
(a) How far does it take to reach the top?
(b)How long does it take to reach the top?
(c)What ir its velocity just before reaching the ground?
V=GT
20=10t
t=2seconds
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I apologize for any confusion. Here are the detailed steps to solve this problem:
Given:
- Initial velocity of the stone, u = 20 m/s
- Acceleration due to gravity, g = 10 m/s^2 (downward direction)
(a) How far does it take to reach the top?
At the top point of its trajectory, the final velocity of the stone (v) becomes zero. We can use the kinematic equation:
v^2 = u^2 + 2gh
where h is the maximum height reached by the stone.
Substituting the given values, we get:
0 = (20)^2 - 2(10)h
h = 20 m
Therefore, it would take the stone a distance of 20 meters to reach the top of its trajectory.
(b) How long does it take to reach the top?
We can use the kinematic equation:
v = u + gt
where v = 0 (at the top), u = 20 m/s, and g = -10 m/s^2 (negative due to upward motion).
Substituting the values, we get:
0 = 20 - 10t
t = 2 seconds
Therefore, it would take the stone 2 seconds to reach the top of its trajectory.
(c) What is its velocity just before reaching the ground?
At the point just before reaching the ground, the height of the stone would be the same as the height at which it was thrown upwards.
Using the same kinematic equation:
v^2 = u^2 + 2gh
we can find the final velocity (v) just before hitting the ground.
Substituting the values, we get:
v^2 = (20)^2 + 2(-10)(20)
v^2 = 400 - 400
v^2 = 0
v = 0 m/s
Therefore, the velocity of the stone just before reaching the ground would be zero.