What angle of elevation θ should you direct the throw a baseball.
a.
Expert Solution
Answer to Problem 110AYU
θ=21.18°,42.36°
Explanation of Solution
Given information:
If you can throw a baseball with an initial speed of 40 meters per second, at what angle of elevation θ should you direct the throw so that the ball travels a distance of 110 meters before striking the ground?
Calculation:
The range of the projectile is given as
R(θ)=v02sin(2θ)g......(1)
We have to find out the angle of elevation such that the ball travels a distance of 110 meters before striking the ground.
Consider the values
v0=40m/s and g=9.8m/s2 .
Let us put these values in equation R(θ)=v02sin(2θ)g......(1) .
R(θ)=v02sin(2θ)g
110=(40)2sin(2θ)9.8
sin(2θ)=(110)(9.8)(40)2
sin(2θ)=0.673
Hence, solving this equation, we get
2θ=42.36°,137.64°
Or
θ=21.18°,42.36°
To determine
Determine the maximum distance that you can throw the ball.
Expert Solution
Answer to Problem 110AYU
Rmax(θ)=163.3meters
Explanation of Solution
Given information:
Determine the maximum distance that you can throw the ball.
Calculation:
The range of the projectile is given as
R(θ)=v02sin(2θ)g......(1)
Throwing maximum distance means maximum range ,that is maximum value of R(θ) in equation (1) . As v0 and g are constant we can only maximize sin(2θ) . And we know that maximum value of sine function is 1. Hence, we have
Rmax(θ)=v02sin(2θ)g
=(40)2(1)9.8
Rmax(θ)=163.3meters
To determine
Plot the graph.
Expert Solution
Answer to Problem 110AYU
Explanation of Solution
Given information:
Graph R=R(θ) , with v0=40 meters per second.
Calculation:
The range of the projectile is given as
R(θ)=v02sin(2θ)g......(1)
Figure below shows the graph of R=R(θ) with v0=40 m/s.
R(θ)=v02sin(2θ)g
To determine
Verify the results obtained in parts (a) and (b).
Expert Solution
Answer to Problem 110AYU
The results in part (a) and (b) are verified.
Explanation of Solution
Given information:
Verify the results obtained in parts (a) and (b) using a graphing utility.
Calculation:
The range of the projectile is given as
R(θ)=v02sin(2θ)g......(1)
We can observe from the graph that range of 110 meters can be achieved at an angle of 21.18° and 42.36° , and maximum range is approximately 163.3 meters.
Here we use TI−83 calculator to graph the function of range and find the maximum distance.
First insert the equation R(θ)=v02sin(2θ)g with v0=40m/s and g=9.8m/s2 that is R(θ)=v02sin(2θ)g
=(40)2sin(2θ)9.8
=163.3sin(2θ)
Now press WINDOW key and set the range of viewing window as.
By pressing the GRAPH key we get
Now for finding the maximum point we use the following process:
Press 2nd key and then TRACE key.
Press 4th key to select MAXIMUM.
Move the cursor to the left of the maximum point, then press ENTER key to select left bound, now move the cursor to the right of the maximum point, then press ENTER key to select right bound now by pressing the ENTER key third time we get the maximum point as
x=0.785,y=163.3
Here we see maximum range is
y=163.3
Also by using the VALUE feature we get that range of 110 meters can be achieved at an angle of 21.18° and 42.36° .
Hence, the results in part (a) and (b) are verified.
Calculus, Single Variable: Early Transcendentals (3rd Edition)
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