A stone thrown upward to just reach the top of a palm tree that is 8.2m tall. Calculate the initial speed that is require
(1/2) v^2 = 9.81 (8.2)
v^2 = 16.4*9.81
160m/sec
more like 12.7
did you take the square root?
I think not :)
I want to learn
To calculate the initial speed required to throw a stone upward to just reach the top of a palm tree, you can use the equations of motion. The key information we need is the final height reached by the stone, which is equal to the height of the palm tree. In this case, the final height is 8.2 meters.
The equation that relates the initial speed (Vi), final speed (Vf), acceleration (a), and displacement (Δy) is:
Vf^2 = Vi^2 + 2aΔy
Since the stone is thrown upward, the final speed at the top of its trajectory is 0 (as it momentarily stops before falling back down). Therefore, we can rewrite the equation as:
0 = Vi^2 + 2aΔy
We also know that the acceleration (a) due to gravity is -9.8 meters per second squared, as gravity acts downward. The displacement (Δy) is +8.2 meters, as the stone moves upward.
Substituting these values into the equation, we get:
0 = Vi^2 + 2(-9.8)(8.2)
Simplifying further:
0 = Vi^2 - 160.76
Rearranging the equation to solve for Vi:
Vi^2 = 160.76
Taking the square root of both sides:
Vi = √160.76
Vi ≈ 12.68 meters per second (rounded to two decimal places)
Therefore, the initial speed required to throw the stone upward to just reach the top of the palm tree is approximately 12.68 meters per second.