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.