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jfarrar02
07.11.2019 •
Mathematics
Aprojectile is fired from ground level at an angle of 12° with the horizontal. the projectile is to have a range of 220 feet. find the minimum initial velocity necessary. (round your answer to one decimal place.)
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Ответ:
The minimum initial velocity has to be 127.8 ft/s
Step-by-step explanation:
Hi there!
The position vector of the projectile can be calculated as follows:
r =(x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).
Please, see the attached figure for a better understanding of the problem. Notice that when the projectile reaches the ground, its position vector will be:
r final = (220, 0) feet
Then, using the equations:
200 feet = x0 + v0 · t · cos α
0 feet = y0 + v0 · t · sin α + 1/2 · g · t²
Since the origin of the frame of reference is located at the launching point, x0 and y0 = 0. Then:
200 ft = v0 · t · cos 12°
0 ft = v0 · t · sin 12° - 1/2 · 32.2 ft/s² · t²
We have a system of equations. Let´s solve the first equation for v0:
200 ft = v0 · t · cos 12°
200 ft/(t · cos 12°) = v0
Now, let´s replace v0 in the second equation:
0 ft = v0 · t · sin 12° - 1/2 · 32.2 ft/s² · t²
0 ft = 200 ft/(t · cos 12°) · t · sin 12° - 16.1 ft/s² · t² (sin 12°/cos 12° = tan 12°)
0 ft = 200 ft · tan 12° - 16.1 ft/s² · t²
(0 ft - 200 ft · tan 12°) / -16.1 ft/s² = t²
t = 1.6 s
Now, let´s calculate v0:
200 ft/(t · cos 12°) = v0
200 ft/(1.6 s · cos 12°) = v0
v0 = 127.8 ft/s
Ответ:
A
Step-by-step explanation:
Parallel lines have equal slopes
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Given
3x + 2y = 6 ( subtract 3x from both sides )
2y = - 3x + 6 ( divide through by 2 )
y = -
x + 2 ← in slope- intercept form
with slope m = -![\frac{3}{2}](/tpl/images/2599/0655/dbf42.png)
Given
y = -
x + 5 ← in slope- intercept form
with slope m = -![\frac{3}{2}](/tpl/images/2599/0655/dbf42.png)
Since the slopes are equal then the lines are parallel