A rock is thrown straight upward with an initial velocity of 9.6 m/s in a location where the acceleration due to gravity ahs a magnitude of 9.81 m/s^2. To what height does it rise?
please walk me through this problem:)
When it stops rising, its speed is zero
V = Vi - 9.81 t
0 = 9.6 - 9.81 t
t = 9.6/9.81
then use
h = ho + Vi t - (9.81/2)t^2
h = 0 + 9.6 (9.6/9.81) - (9.81/2)(9.6/9.81)^2
There is also a shortcut way to do this using "energy conservation" that will give the same answer. At the maximum height H, the potential energy increase M g H will equal the initial kinetic energy, (1/2) M Vo^2)
Therefore H = Vo^2/(2g)
Sure! Let's work through this problem step-by-step.
Step 1: Identify the given information.
The problem states that the rock is thrown upwards with an initial velocity of 9.6 m/s and the acceleration due to gravity is 9.81 m/s^2.
Step 2: Determine the variables needed to solve the problem.
We need to find the height to which the rock rises. Let's call this variable "h" for height.
Step 3: Determine the equations to use.
We can use the following kinematic equation to solve for the height:
vf^2 = vi^2 + 2ad
where
vf = final velocity (which will be 0 when the rock reaches its maximum height)
vi = initial velocity
d = displacement (height in this case)
a = acceleration due to gravity
Step 4: Plug in the known values into the equation.
vf^2 = 0 (since the velocity at the top is 0)
vi = 9.6 m/s
a = 9.81 m/s^2
0 = (9.6 m/s)^2 + 2(9.81 m/s^2)(h)
Simplifying this equation, we have:
0 = 92.16 m^2/s^2 + 19.62 m/s^2(h)
Step 5: Solve for the height (h).
To isolate h, we subtract 92.16 m^2/s^2 from both sides of the equation:
-92.16 m^2/s^2 = 19.62 m/s^2(h)
Divide both sides by 19.62 m/s^2:
-4.7 = h
Step 6: Analyze the result.
The value of h is negative, indicating that the height is below the starting point. Therefore, the rock actually falls to a height of 4.7 meters below the starting point.
So, the height to which the rock rises is 4.7 meters below the initial position.
Certainly! To find the height to which the rock rises, we can use the kinematic equation for vertical motion:
h = (v^2 - u^2) / (2a)
Where:
h = height
v = final velocity
u = initial velocity
a = acceleration due to gravity
In this case, the rock is thrown straight upward, so its final velocity at the highest point will be 0 m/s. We are given that the initial velocity u is 9.6 m/s, and the acceleration due to gravity a is 9.81 m/s^2.
Now, let's substitute these values into the equation:
h = (0^2 - 9.6^2) / (2 * -9.81)
= (-92.16) / (-19.62)
≈ 4.7 meters (rounded to one decimal place)
Therefore, the rock rises to a height of approximately 4.7 meters.
Note: The negative sign in the acceleration term is because the acceleration due to gravity points downward, while the positive direction is taken as upward. This means that we assign negative values to the downward direction and positive values to the upward direction.