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?
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.
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
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.
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.