a boy throws a stone straight upward with an initial speed of 15.0m/s. what mazimum height will the stone reach before falling back down.
Can you use V^2 = Vo^2 + 2a*distance and set V^2 = 0?
One method to solve this is conservation of energy.
initial kinetic energy = final potential energy
let m = mass of stone
let g = acceleration of gravity
let v = initial velocity
let h = maximum height
(1/2)mv^2 = mgh
So,
v^2 = 2gh
h = (v^2)/g
23
22.95
final velocity squared = initial velocity squared + 2(acceleration)(displacement)
0 = (15*15) + 2(9.8)x
0 = 225 + 19.6x
-225 = 19.6x
x= 11m
To find the maximum height reached by the stone, we can use the equations of motion. Let's break down the problem step by step:
Step 1: Identify the given information:
The initial speed of the stone, u = 15.0 m/s.
Step 2: Determine what information is needed:
We need to find the maximum height (h) reached by the stone.
Step 3: Find the necessary equation(s):
We can use the equation for displacement (change in height) in vertical motion:
Δh = v^2 - u^2 / 2g
Where:
Δh = change in height (maximum height - initial height)
v = final velocity (0 m/s at maximum height, as the stone turns around)
u = initial velocity (15.0 m/s, given)
g = acceleration due to gravity (approximately 9.8 m/s^2)
Step 4: Plug in the values and calculate:
As the stone reaches its maximum height, the final velocity (v) is 0 m/s. Plugging in our known values:
Δh = 0^2 - 15^2 / (2 * 9.8)
Δh = 225 / 19.6
Δh ≈ 11.48 m
Therefore, the maximum height reached by the stone before falling back down is approximately 11.48 meters.