Use g 9.9 m/s and an initial velocity of 100 m/s. Show that the maximum range is obtained at 0 = 7/4 by computing and plotting the range for values of O from 0ses in increments of 0.05. Repeat your calculations with an initial velocity of 50 m/s, and plot both sets of results on a single graph. 2/4

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BALLISTICS
The range of an object (see Figure 5.12) shot at an angle 0 with respect to the xaxis
and an initial velocity th is given by
R(0) =
sin (20)
for 0 s0s(neglecting air resistance)
Use g= 9.9 m/s and an initial velocity of 100 m/s. Show that the maximum
range is obtained at 0 = 7/4 by computing and plotting the range for values of
O from
in increments of 0.05.
Repeat your calculations with an initial velocity of 50 m/s, and plot both sets of
results on a single graph.
1. State the Problem
Calculate the range as a function of the launch angle.
2. Describe the Input and Output
Input
g= 9.9 m/s
0 = 0 to 7/2, incremented by 0.05
50 m/s and 100 m/s
%3D
Transcribed Image Text:BALLISTICS The range of an object (see Figure 5.12) shot at an angle 0 with respect to the xaxis and an initial velocity th is given by R(0) = sin (20) for 0 s0s(neglecting air resistance) Use g= 9.9 m/s and an initial velocity of 100 m/s. Show that the maximum range is obtained at 0 = 7/4 by computing and plotting the range for values of O from in increments of 0.05. Repeat your calculations with an initial velocity of 50 m/s, and plot both sets of results on a single graph. 1. State the Problem Calculate the range as a function of the launch angle. 2. Describe the Input and Output Input g= 9.9 m/s 0 = 0 to 7/2, incremented by 0.05 50 m/s and 100 m/s %3D
Figure 5.12
The ronge is zero, i the conon is perfecty vertical or perfecty horizontal.
Output
Range R
Present the results as a plot.
3. Develop a Hand Example
If the cannon is pointed straight up, we know that the range is zero, and if the
cannon is horizontal, the range is also zero (see Figure 5.12).
This means that the range must increase with the cannon angle up to some
maximum and then decrease. A sample calculation at 45 (/4 radians) shows
that
- sin (20)
%3D
100
sin
9.9
(等)
= 1010 m when the initial velocity is 100 m/s
Transcribed Image Text:Figure 5.12 The ronge is zero, i the conon is perfecty vertical or perfecty horizontal. Output Range R Present the results as a plot. 3. Develop a Hand Example If the cannon is pointed straight up, we know that the range is zero, and if the cannon is horizontal, the range is also zero (see Figure 5.12). This means that the range must increase with the cannon angle up to some maximum and then decrease. A sample calculation at 45 (/4 radians) shows that - sin (20) %3D 100 sin 9.9 (等) = 1010 m when the initial velocity is 100 m/s
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