Question
in order to save time one worker decides to throw his apple up to his co-worker he releases the apple from a height of 2.00m above the ground level with an initial speed of 15.5m/s what is the maximum height of the apple
Answers
Answered by
GPT 3.5
To find the maximum height of the apple, we can use the equations of motion.
The initial height of the apple is 2.00m and its initial speed is 15.5m/s. We need to determine the maximum height it reaches before falling back down.
The equation we can use is:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s at maximum height, as it momentarily stops before falling back down)
u = initial velocity (15.5 m/s)
a = acceleration (due to gravity, which is -9.8 m/s^2)
s = displacement (maximum height reached)
Rearranging the equation, we have:
0^2 = (15.5)^2 + 2(-9.8)s
0 = 240.25 - 19.6s
19.6s = 240.25
s = 240.25 / 19.6
s ≈ 12.27m
Therefore, the maximum height the apple reaches is approximately 12.27m.
The initial height of the apple is 2.00m and its initial speed is 15.5m/s. We need to determine the maximum height it reaches before falling back down.
The equation we can use is:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s at maximum height, as it momentarily stops before falling back down)
u = initial velocity (15.5 m/s)
a = acceleration (due to gravity, which is -9.8 m/s^2)
s = displacement (maximum height reached)
Rearranging the equation, we have:
0^2 = (15.5)^2 + 2(-9.8)s
0 = 240.25 - 19.6s
19.6s = 240.25
s = 240.25 / 19.6
s ≈ 12.27m
Therefore, the maximum height the apple reaches is approximately 12.27m.