A stone is thrown vertically upward from the ground with an initial velocity of 20m/s.(i)What is the maximum height reached ?,(ii)How long does the ball take to reach this maximum height?

(i) To find the maximum height reached, we can use the kinematic equation:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (20 m/s)
a = acceleration due to gravity (-9.8 m/s^2, negative because it is acting in the opposite direction of the motion)
s = displacement (maximum height)

Rearranging the equation to solve for s:

0 = (20)^2 + 2(-9.8)s
0 = 400 - 19.6s
19.6s = 400
s = 400/19.6
s ≈ 20.41 m

Therefore, the maximum height reached by the stone is approximately 20.41 meters.

(ii) To find the time taken to reach the maximum height, we can use the kinematic equation:

v = u + at

Where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (20 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time taken to reach maximum height

Rearranging the equation to solve for t:

0 = 20 - 9.8t
9.8t = 20
t = 20/9.8
t ≈ 2.04 seconds

Therefore, the stone takes approximately 2.04 seconds to reach the maximum height.

(i) Maximum Height Reached

Using the equation of motion:

```
v^2 = u^2 + 2as
```

where:

- v is the final velocity (0 m/s at the maximum height)
- u is the initial velocity (20 m/s)
- a is the acceleration due to gravity (-9.8 m/s²)
- s is the maximum height reached

Substituting the values:

```
0^2 = 20^2 + 2(-9.8)s
s = 20.41 m
```

Therefore, the maximum height reached is 20.41 m.

(ii) Time Taken to Reach Maximum Height

Using the equation of motion:

```
v = u + at
```

where:

- v is the final velocity (0 m/s at the maximum height)
- u is the initial velocity (20 m/s)
- a is the acceleration due to gravity (-9.8 m/s²)
- t is the time taken

At the maximum height, the final velocity is zero. Substituting the values:

```
0 = 20 - 9.8t
t = 2.04 s
```

Therefore, it takes 2.04 seconds to reach the maximum height.