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anniekwilbourne
15.04.2020 •
Physics
An object is thrown with velocity v from the edge of a cliff above level ground. Neglect air resistance. In order for the object to travel a maximum horizontal distance from the cliff before hitting the ground, the throw should be at an angle θ with respect to the horizontal of:.
(A) greater than 60° above the horizontal
(B) greater than 45° but less than 60° above the horizontal
(C) greater than zero but less than 45° above the horizontal
(D) zero
(E) greater than zero but less than 45° below the horizontal
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Ответ:
The correct option is;(C) greater than zero but less than 45° above the horizontal
Explanation:
Here, we have for maximum horizontal distance
h = v×t
Where t is the time of flight
The time of flight is given by
s = v·t - 0.5 × gt² to maximize the time of flight, we therefore increase the height such that
Since the range is given by
Horizontal range, x = v·t·cosα
Vertical range, y = v·t·sinα - 0.5·g·t²
When the particle comes back to initial level, we have
0 = v·t·sinα - 0.5·g·t² → 0 = t(v·sinα - 0.5·g·t)
So that t = 0 or t =![\frac{2\cdot v\cdot sin\alpha }{g}](/tpl/images/0601/1033/1e6e0.png)
Therefore, horizontal range =
Therefore maximum range is obtained when α = 45° as sin 90° = 1
Ответ:
A:)
Step-by-step explanation: