Suppose that an object is dropped from a height of h meters and hits the ground with a velocity of v meters per second. Then v= sqrt19.6h. If an object hits the ground with a velocity of 18.5 meters per second, from what height was it dropped?
Carry your intermediate computations to at least four decimal places, and round your answer to the nearest tenth.
We are given the equation relating the velocity (v) and the height (h) of an object when it hits the ground: v = sqrt(19.6h).
We are also given that the velocity of the object is 18.5 m/s. Using this information, we can substitute v = 18.5 into the equation to find the height (h):
18.5 = sqrt(19.6h)
To solve for h, we need to isolate it. Start by squaring both sides of the equation:
(18.5)^2 = (sqrt(19.6h))^2
342.25 = 19.6h
Next, divide both sides of the equation by 19.6:
342.25 / 19.6 = h
17.4627551 = h
Rounding to the nearest tenth, the object was dropped from a height of approximately 17.5 meters.
To find the height from which the object was dropped, we can rearrange the given equation v = sqrt(19.6h) to solve for h.
v = sqrt(19.6h)
First, square both sides of the equation:
v^2 = 19.6h
Substitute the given velocity into the equation:
(18.5 m/s)^2 = 19.6h
Simplify the equation:
342.25 m^2/s^2 = 19.6h
Now, divide both sides of the equation by 19.6:
342.25 m^2/s^2 / 19.6 = h
Solve for h:
h ≈ 17.4423 meters
Therefore, the object was dropped from a height of approximately 17.4 meters.
Come on -- they gave you the formula!
√(19.6h) = 18.5
19.6h = 18.5^2
h = 18.5^2/19.6