If a stone is thrown upward at 29.7 m/s, what will be its velocity at a height of 12 meters?
Vf^2 = Vo^2 + 2gd,
Vf^2 = (29.7)^2 + 2(-9.8)*12 = 646.89,
Vf = 25.4m/s.
To determine the velocity of the stone at a height of 12 meters, you need to consider the principles of motion and apply the relevant formulas.
When the stone is thrown upward, it experiences the force of gravity pulling it downwards, causing it to decelerate. At some point, the stone will reach its highest point (the apex) and then start to descend.
To find the velocity of the stone at a specific height, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity (throwing velocity)
a = acceleration due to gravity (approximately -9.8 m/s^2)
s = displacement (change in height)
Given:
u = 29.7 m/s (upwards)
s = 12 m
a = -9.8 m/s^2
Now, let's plug in these values into the equation:
v^2 = (29.7 m/s)^2 + 2(-9.8 m/s^2)(12 m)
v^2 = 882.09 m^2/s^2 + (-235.2 m^2/s^2)
v^2 = 646.89 m^2/s^2 - 235.2 m^2/s^2
v^2 = 411.69 m^2/s^2
To obtain the velocity, take the square root of both sides:
v = sqrt(411.69 m^2/s^2)
v ≈ 20.29 m/s
Therefore, the approximate velocity of the stone at a height of 12 meters is 20.29 m/s.