A BALL IS PROJECTED VERTICALLY UPWARD FROM THE TOP OF A TOWER 60MITRE HIGH WITH A VELOCITY OF 30MITRE PER SECONDS.WHAT IS THE MAXIMUM HEIGHT ABOVE THE GROUND LEVEL?HOW LONG DOES IT TAKE TO REACH THE GROUND?

a. h = ho + (Vf^2-Vo^2)/2g.

ho = 60 m.
Vf = 0.
Vo = 30 m/s.
g = -9.8 m/s^2.
h = ?

b. Vf = Vo + g*Tr.
Vf = 0.
Vo = 30 m/s.
g = -9.8 m/s^2.
Tr = Rise time.
Tr = ?

b. 0.5g*Tf^2 = h(Part a).
g = 9.8 m/s^2.
Tf = Fall time.
Tf = ?

Tr + Tf = Time to reach gnd.

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To find the maximum height above the ground level and the time it takes for the ball to reach the ground, we can use the equations of motion.

First, let's find the maximum height above the ground level:

1. Identify the given values:
- Initial velocity (u) = 30 m/s (upward direction)
- Acceleration (a) = -9.8 m/s^2 (downward direction, due to gravity)
- Displacement (s) = ?
- Final velocity (v) = 0 m/s (at maximum height)
- Time (t) = ?

2. Use the equation for displacement:
s = ut + (1/2)at^2

Since the ball is projected vertically upward and comes to rest at maximum height, the final velocity (v) is 0 m/s. Thus, we can rearrange the equation and solve for displacement (s):
0 = 30t + (1/2)(-9.8)t^2

3. Simplify the equation:
-4.9t^2 + 30t = 0

4. Factor out t:
t(-4.9t + 30) = 0

This equation suggests that there are two potential solutions: t = 0 or -4.9t + 30 = 0

5. Solve for t:
-4.9t + 30 = 0
-4.9t = -30
t = -30 / -4.9
t ≈ 6.12 seconds

Since time cannot be negative in this context, we discard t = 0 and consider t ≈ 6.12 seconds as the time taken to reach maximum height.

6. Calculate the maximum height:
s = ut + (1/2)at^2
s = 30(6.12) + (1/2)(-9.8)(6.12)^2
s ≈ 183.6 - 181.3472
s ≈ 2.25 meters

Therefore, the maximum height above the ground level is approximately 2.25 meters.

To find out how long it takes for the ball to reach the ground, we can use the equation for time of flight:

1. Identify the given values:
- Initial velocity (u) = 30 m/s (upward direction)
- Acceleration (a) = -9.8 m/s^2 (downward direction, due to gravity)
- Displacement (s) = -60 m (negative since the ball is coming down)
- Final velocity (v) = ?
- Time (t) = ?

2. Use the equation for displacement:
s = ut + (1/2)at^2

Since the ball is coming down to the ground level, the final displacement (s) is -60 m. Thus, we can rearrange the equation and solve for time (t):
-60 = 30t + (1/2)(-9.8)t^2

3. Simplify the equation:
-4.9t^2 + 30t + 60 = 0

4. Solve for t using the quadratic formula (or factoring if possible):
t = (-b ± √(b^2 - 4ac))/(2a)
where a = -4.9, b = 30, and c = 60

t = (-30 ± √(30^2 - 4(-4.9)(60)))/(2(-4.9))

Solve for t using the quadratic formula to get two potential solutions: t1 and t2.

5. The positive solution will give us the time taken to reach the ground:
t ≈ 6.17 seconds

Therefore, it takes approximately 6.17 seconds for the ball to reach the ground.