For Problems 121-124, use the following discussion. Projectile Motion The path of a projectile fired at an inclination θ to the horizontal with initial speed v 0 is a parabola (see the figure). The range R of the projectile, that is, the horizontal distance that the projectile travels, is found by using the function R ( θ ) = v 0 2 sin ( 2 θ ) g where g ≈ 32.2 feet per second per second ≈ 9.8 meters per second per second is the acceleration due to gravity. The maximum height H of the projectile is given by the function H ( θ ) = v 0 2 ( sin θ ) 2 g 2 In Problems 121-124, find the range R and maximum height H . (See the discussion on the previous page.) The projectile is fired at an angle of 30 ∘ to the horizontal with an initial speed of 150 meters per second.
For Problems 121-124, use the following discussion. Projectile Motion The path of a projectile fired at an inclination θ to the horizontal with initial speed v 0 is a parabola (see the figure). The range R of the projectile, that is, the horizontal distance that the projectile travels, is found by using the function R ( θ ) = v 0 2 sin ( 2 θ ) g where g ≈ 32.2 feet per second per second ≈ 9.8 meters per second per second is the acceleration due to gravity. The maximum height H of the projectile is given by the function H ( θ ) = v 0 2 ( sin θ ) 2 g 2 In Problems 121-124, find the range R and maximum height H . (See the discussion on the previous page.) The projectile is fired at an angle of 30 ∘ to the horizontal with an initial speed of 150 meters per second.
Solution Summary: The author explains how to find the range R and maximum height H.
For Problems 121-124, use the following discussion.
Projectile Motion The path of a projectile fired at an inclination
to the horizontal with initial speed
is a parabola (see the figure).
The range
of the projectile, that is, the horizontal distance that the projectile travels, is found by using the function
where
feet per second per second
meters per second per second is the acceleration due to gravity. The maximum height
of the projectile is given by the function
In Problems 121-124, find the range
and maximum height
. (See the discussion on the previous page.)
The projectile is fired at an angle of
to the horizontal with an initial speed of 150 meters per second.
The cup on the 9th hole of a golf course is located dead center in the middle of a circular green which is 40 feet in radius. Your ball is located as in the picture below. The ball follows a straight line path and exits the green at the right-most edge. Assume the ball travels 8 ft/sec.
Introduce coordinates so that the cup is the origin of an xy-coordinate system and start by writing down the equations of the circle and the linear path of the ball. Provide numerical answers below with two decimal places of accuracy.
50 feet
green
ball
40 feet
9
cup
ball path
rough
(a) The x-coordinate of the position where the ball enters the green will be
(b) The ball will exit the green exactly
seconds after it is hit.
(c) Suppose that L is a line tangent to the boundary of the golf green and parallel to the path of the ball. Let Q be the point where the line is tangent to the circle. Notice that there are two possible positions for Q. Find the possible x-coordinates of Q:
smallest x-coordinate =…
Draw the unit circle and plot the point P=(8,2). Observe there are TWO lines tangent to the circle passing through the point P. Answer the questions below with 3 decimal places of accuracy.
P
L1
L
(a) The line L₁ is tangent to the unit circle at the point
(b) The tangent line L₁ has equation:
X +
(c) The line L₂ is tangent to the unit circle at the point (
(d) The tangent line 42 has equation:
y=
x +
).
What is a solution to a differential equation? We said that a differential equation is an equation that
describes the derivative, or derivatives, of a function that is unknown to us. By a solution to a differential
equation, we mean simply a function that satisfies this description.
2. Here is a differential equation which describes an unknown position function s(t):
ds
dt
318
4t+1,
ds
(a) To check that s(t) = 2t2 + t is a solution to this differential equation, calculate
you really do get 4t +1.
and check that
dt'
(b) Is s(t) = 2t2 +++ 4 also a solution to this differential equation?
(c) Is s(t)=2t2 + 3t also a solution to this differential equation?
ds
1
dt
(d) To find all possible solutions, start with the differential equation = 4t + 1, then move dt to the
right side of the equation by multiplying, and then integrate both sides. What do you get?
(e) Does this differential equation have a unique solution, or an infinite family of solutions?
Chapter 6 Solutions
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