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 =v19.6h. If an object hits the ground with a velocity of 27.7 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.
(1/2) m v^2 =m g h
on earth g is about 9.8m/s^2
v^2 = 2 g h
v = sqrt (2 g h) = sqrt (2*9.8*h) = sqrt ( 19.6 h) ok so far
now
27.7 = sqrt (19.6 h)
27.7^2 = 19.6 h
767/19.6 = h
To find the height from which the object was dropped, we can use the formula v^2 = u^2 + 2gh, where v is the final velocity (27.7 m/s), u is the initial velocity (0 m/s), g is the acceleration due to gravity (9.8 m/s^2), and h is the height.
Rearranging the formula, we get h = (v^2 - u^2) / (2g).
Substituting the given values, we have h = (27.7^2 - 0^2) / (2 * 9.8).
Calculating the numerator, we have h = (767.29 - 0) / (2 * 9.8).
Simplifying further, we get h = 767.29 / 19.6.
Calculating the division, we find h = 39.168367346938775.
Rounding to the nearest tenth, the height from which the object was dropped is approximately 39.2 meters.
To solve this problem, we can use the equation =v^2-2gh, where is the final velocity, v is the initial velocity (in this case, 0 m/s), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Given that the object hits the ground with a velocity of 27.7 m/s, we can substitute these values into the equation:
27.7^2 = 0 - 2(9.8)h
Simplifying,
766.29 = -19.6h
Dividing both sides by -19.6,
h = 766.29 / -19.6
Calculating this value, we get:
h ≈ -39.14
Since height cannot be negative in this context, we round the answer to the nearest tenth:
h ≈ 39.1 meters
Therefore, the object was dropped from a height of approximately 39.1 meters.