Mathematics: 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?

Thomas (t) = Andrew (x) * 2 + 12Step-by-step explanation:...

Recall that the equation for horizontal distance

marisol84

17.01.2022

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

$h=\frac{v²_{0}}{16}sin\theta cos\theta$

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

Then the equation will be $h=\frac{v_{0}^{2}}{16}sin\theta cos\theta$

$100=\frac{60^{2}}{32}sin2\theta$

$100=\frac{3600}{32}sin2\theta$

$sin2\theta =32\times 100/3600=0.89$

$2\theta =62.87$

$\theta =31.43$

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

$h=\frac{60^{2}}{32}sin2\times 45$

$=\frac{3600}{32}sin90$

$=112.5\times 1=225$

h = 112.5 feet

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