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/s2. To what height does it rise? Help meh please.
d = v^2/(2a)
d = 9.6^2/(2(9.81))
d = 4.7 m haha i answered it myself
so confused
To determine the height the rock rises, we can use the equation of motion for an object in free fall:
h = (v^2 - u^2) / (2a)
where:
h = height
v = final velocity (in this case, at the top of the motion, the velocity will be 0 m/s)
u = initial velocity
a = acceleration due to gravity
Let's substitute the given values into the equation:
h = (0^2 - 9.6^2) / (2 * -9.81)
First, let's simplify the equation:
h = (-92.16) / (-19.62)
Dividing both sides by -1 to cancel out the negative sign:
h = 92.16 / 19.62
Calculating the value:
h ≈ 4.70 meters
Therefore, the rock rises to a height of approximately 4.70 meters.
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