A stone is thrown straight up with a velocity of 24.0m/s from a height of 2.00m. How fast is it moving when it is at a height of 13.0m? How much time is required to reach this height?
13 = -½ g t² + 24 t + 2
solve for t...there will be two solutions (up and down)
the velocity is...v = 24 - g t
...use the up (smaller) time
a. V^2 = Vo^2 + 2g*h = 24^2 - 19.6 *(13-2) =
b. V = Vo + g*t. V = (part a), Vo = 24 m/s, g = -9.8m/s^2, t = ?.
To solve this problem, we can use the laws of motion and principles of kinematics. Let's break it down into steps:
Step 1: Determine the initial velocity and initial height:
Given:
Initial velocity, u = 24.0 m/s
Initial height, h = 2.00 m
Step 2: Calculate the time taken to reach the height of 13.0m:
Using the kinematic equation:
h = ut + (1/2)gt^2
where:
h = height
u = initial velocity
g = acceleration due to gravity (approximately 9.8 m/s²)
t = time taken
Rearranging the equation, we get:
t = (sqrt(2h/g))
Substituting the values:
t = (sqrt(2 * 13.0 / 9.8))
= (sqrt(26 / 9.8))
= (sqrt(2.653))
≈ 1.63 seconds
So, it will take approximately 1.63 seconds to reach a height of 13.0m.
Step 3: Determine the velocity when it reaches the height of 13.0m:
Using the formula:
v = u + gt
Where:
v = final velocity
u = initial velocity
g = acceleration due to gravity
t = time taken
Substituting the values:
v = 24.0 + (-9.8 * 1.63)
= 24.0 - 15.974
≈ 8.03 m/s
Therefore, the stone will be moving at approximately 8.03 m/s when it is at a height of 13.0 m.
To solve this problem, we can use the equations of motion for an object thrown vertically.
First, let's find the initial velocity (u), final velocity (v), initial height (h1), and final height (h2).
Given:
Initial velocity (u) = 24.0 m/s (positive because it's going up)
Initial height (h1) = 2.00 m
Final height (h2) = 13.0 m
The acceleration due to gravity (g) is a constant of -9.8 m/s^2 (negative because it acts in the opposite direction to the motion).
Next, let's use the equations of motion:
1. vf = u + at
2. h2 = h1 + ut + (1/2)at^2
3. vf^2 = u^2 + 2aΔh
For equation 1, we have:
vf = final velocity (unknown)
u = initial velocity = 24.0 m/s
a = acceleration due to gravity = -9.8 m/s^2 (negative because it acts in the opposite direction)
t = time (unknown)
Using equation 1, we can solve for vf:
vf = u + at
vf = 24.0 m/s - 9.8 m/s^2 * t
Now let's use equation 2 to solve for t:
h2 = h1 + ut + (1/2)at^2
13.0 m = 2.00 m + (24.0 m/s) * t + (0.5 * -9.8 m/s^2) * t^2
Simplifying equation 2:
0.5 * -9.8 m/s^2 * t^2 + 24.0 m/s * t + 2.00 m - 13.0 m = 0
Now we can solve this quadratic equation to find t.