Asked by Christopher.M.C.Brown
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?
(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?
Answers
Answered by
oobleck
v = 20 - 9.8t
the stone is at the top when v has gone to zero
h = 20t - 4.9t^2
Use t found above to get h
since energy is conserved, its speed on return is the same as its speed on launch.
the stone is at the top when v has gone to zero
h = 20t - 4.9t^2
Use t found above to get h
since energy is conserved, its speed on return is the same as its speed on launch.
Answered by
Japhet Manda
Answer
Answered by
Nelson Mubita
I want more steps on this question
Answered by
Peter
Great it's really helping me
Answered by
Falesi banda
Good
Answered by
Anonymous
Y r there no answers
Answered by
Billy
V=GT
20=10t
t=2seconds
20=10t
t=2seconds
Answered by
Simon
Help me a answer
Answered by
Mirriam
I don't know the answer
Answered by
Bot
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.
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.
Answered by
Thabo
A stone is thrown upwards with an initial velocity of 20m/s air resistance is ignored g=10m/s², how long does it take to reach the ground
Answered by
Bot
When the stone hits the ground, its height (h) would be zero. We can use the same kinematic equation:
h = ut + (1/2)gt^2
where u = 20 m/s (upward), g = -10 m/s^2 (downward), and h = 0.
Substituting the values, we get:
0 = 20t - 5t^2
t(5t - 20) = 0
t = 0 or t = 4 seconds
Since time cannot be negative, the time taken by the stone to reach the ground would be 4 seconds.
Therefore, it would take the stone 4 seconds to reach the ground.
h = ut + (1/2)gt^2
where u = 20 m/s (upward), g = -10 m/s^2 (downward), and h = 0.
Substituting the values, we get:
0 = 20t - 5t^2
t(5t - 20) = 0
t = 0 or t = 4 seconds
Since time cannot be negative, the time taken by the stone to reach the ground would be 4 seconds.
Therefore, it would take the stone 4 seconds to reach the ground.