Recall that the equation for horizontal distance "h" in feet of a projectile with initial velocity v0 and initial angle theta is given by h=(v0^2/16)sin theta cos theta. a.) Assume the initial velocity is 60 (feet/second). What initial angle will you need to ensure that the horizontal distance will be exactly 100 feet? b.) Assume the initial velocity is 60 feet/second. What is the maximum horizontal distance possible and at what angle does this occur?

A

100 = (60^2 / 16) sin theta cos theta
100 = 225 sin theta cos theta
sin theta cos theta = 100/224 = 0.4444
1/2 sin 2theta = 0.4444
sin 2theta = 0.8888
2 * theta = 62.73 degrees
theta = 31.37 degrees

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eudora eudora · Answer 2 · 2019-03-24T04:07:39+01:00

Answer:

A). 31.43°

B). 225 feet

Step-by-step explanation:

A) From the given equation for horizontal distance h of a projectile is given by

[tex]h=\frac{v²_{0}}{16}sin\theta cos\theta[/tex]

If initial velocity v0 = 60 ft/sec and horizontal distance = 100 feet

Then the equation will be [tex]h=\frac{v_{0}^{2}}{16}sin\theta cos\theta[/tex]

[tex]100=\frac{60^{2}}{32}sin2\theta[/tex]

[tex]100=\frac{3600}{32}sin2\theta[/tex]

[tex]sin2\theta =32\times 100/3600=0.89[/tex]

[tex]2\theta =62.87[/tex]

[tex]\theta =31.43[/tex]

B). If v0 = 60 feet per second

Then we have to calculate the maximum horizontal distance and angle.

Since we know to cover the maximum distance in a projectile motion an object should be thrown at 45°.

Therefore the equation formed will be

[tex]h=\frac{60^{2}}{32}sin2\times 45[/tex]

[tex]=\frac{3600}{32}sin90[/tex]

[tex]=112.5\times 1=225[/tex]

h = 112.5 feet