A rock is thrown straight upward with an initial velocity of 9.6 m/s in a location where the acceleration due to gravity has a magnitude of 9.81 m/s^2. To what height does it rise?
i got 4.7
can you please show me how to solve for this last one
The rock rises up to a height H where the velocity is zero, at which point the increase in potential energy M g H equals the initial kinetic energy, (1/2) M V^2
Therefore H = V^2/(2g) = 4.7 m
Since that is the answer you got, you must already know how to do the problem.
Don't forget to show the dimension (meters in this case). In physics and engineering, a number without a dimension should be marked wrong.
To solve this problem, you can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity (when the rock reaches its highest point, vf is 0 m/s)
- vi is the initial velocity (9.6 m/s, as given in the problem)
- a is the acceleration due to gravity (-9.81 m/s^2, as given in the problem)
- d is the displacement (we need to find this)
Since we want to find the height the rock rises to, the displacement is equal to the positive height above the starting point. Therefore, we can rewrite the equation as:
0^2 = (9.6)^2 + 2(-9.81)d
Simplifying the equation:
0 = 92.16 - 19.62d
Rearranging the equation to solve for d:
19.62d = 92.16
d = 92.16 / 19.62
Using a calculator to evaluate:
d ≈ 4.700 meters
Therefore, the height to which the rock rises is approximately 4.700 meters.