a stone is thrown vertically upward at a speed of 10 m/s from a cliff that is 95m high. How long will it take the stone to reach the bottom?

find t when h=0.

h(t) = 95+10t-4.9t^2

To solve this problem, we can use the kinematic equations of motion.

First, let's determine the initial velocity and the acceleration due to gravity.

Given:
Initial velocity (u) = 10 m/s (upward)
Height (h) = 95 m

Acceleration due to gravity (a) = -9.8 m/s² (as it acts downward)

We can use the following equation to determine the time taken (t) to reach the bottom:

h = ut + (1/2)at²

In this case, the final height (h) is the bottom of the cliff, which is 0 m. Therefore, we have:

0 = 10t + (1/2)(-9.8)t²

Simplifying this equation, we get:

0 = 10t - 4.9t²

Rearranging the equation, we have:

4.9t² - 10t = 0

Now, we need to solve this quadratic equation to find the time (t) when the stone reaches the bottom. We can factor out t:

t(4.9t - 10) = 0

This gives us two possible solutions:

1. t = 0 (which is the initial time when the stone is thrown)
2. 4.9t - 10 = 0

Solving the second equation, we get:

4.9t = 10
t = 10 / 4.9
t ≈ 2.04 seconds

Therefore, it will take approximately 2.04 seconds for the stone to reach the bottom of the cliff.