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
            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
            
    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
            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.
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