An object is launched with an initial speed of 30 m/s at an angle of 60° above the horizontal . What is the maximum height reach
1 answer:
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Answer:
Explanation:
For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).
With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:
γhxdy
we know that γ=62.5 lb/
from the drawing, we can determine the height (or depth under the water) of each differential area is given by:
h=8-y
x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:
when substituting the x and y-values given on the graph, we get that the slope is:
once we got this slope, we can substitute it in the point-slope form of the equation:
which yields:
which simplifies to:
we can now solve this equation for x, so we get that:
with this last equation, we can substitute everything into our integral, so it will now look like this:
Now that it's all written in terms of y we can now simplify it, so we get:
we can now proceed and evaluate it.
When using the power rule on each of the terms, we get the integral to be:
By using the fundamental theorem of calculus we get:
When solving we get:
a) 50.8 mm
b) The whole image (1:1)
c) It seems reasonable
Explanation:
a)
To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:
where
f is the focal length of the lens
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem:
f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)
p = 3.00 m is the distance of the person from the lens
Therefore, we can find q:
b)
Here we need to find the height of the image first.
This can be done by using the magnification equation:
where:
y' is the height of the image
y = 1.75 m is the height of the real person
q = 50.8 mm = 0.0508 m is the distance of the image from the lens
p = 3.00 m is the distance of the person from the lens
Solving for y', we find:
(the negative sign means the image is inverted)
Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.
c)
This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.