A stone is thrown upward from the top of 59.4 m high cliff with an upward velocity component of 19.6 ms-¹ .How long the stone will take to reach maximum height?

To find the time it takes for the stone to reach maximum height, we can use the following kinematic equation:

v_f = v_i + at

Where:
v_f = final velocity (0 m/s at maximum height)
v_i = initial velocity (19.6 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time

Rearranging the equation to solve for time (t), we get:

0 = 19.6 - 9.8t
9.8t = 19.6
t = 19.6 / 9.8
t = 2 seconds

Therefore, the stone will take 2 seconds to reach maximum height.

1. Identify the givens:

- Initial height: h₀ = 59.4 m
- Initial upward velocity: v₀ = 19.6 m/s
- Gravitational acceleration: g = 9.8 m/s²

2. Determine the acceleration due to gravity acting downward:

- a = -g = -9.8 m/s²

3. Use the equation of motion for constant acceleration to find the time (t) it takes for the stone to reach maximum height:

- v = v₀ + at
- At maximum height, the velocity (v) becomes zero. So, we have:
- 0 = 19.6 m/s - 9.8 m/s² - t

4. Solve for the time (t):

- Dividing both sides by -9.8 m/s², we get:
- t = 19.6 m/s / 9.8 m/s²
- t = 2.0 s

Therefore, the stone will take 2.0 seconds to reach maximum height.