Aprojectile is fired from ground level at an angle of 12° with the horizontal. the projectile is to have a range of 220 feet. find the minimum initial velocity necessary. (round your answer to one decimal place.)

The minimum initial velocity has to be 127.8 ft/s

Step-by-step explanation:

Hi there!

The position vector of the projectile can be calculated as follows:

r =(x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).

Please, see the attached figure for a better understanding of the problem. Notice that when the projectile reaches the ground, its position vector will be:

r final = (220, 0) feet

Then, using the equations:

200 feet = x0 + v0 · t · cos α

0 feet = y0 + v0 · t · sin α + 1/2 · g · t²

Since the origin of the frame of reference is located at the launching point, x0 and y0 = 0. Then:

200 ft = v0 · t · cos 12°

0 ft =  v0 · t · sin 12° - 1/2 · 32.2 ft/s² · t²

We have a system of equations. Let´s solve the first equation for v0:

200 ft = v0 · t · cos 12°

200 ft/(t · cos 12°) = v0

Now, let´s replace v0 in the second equation:

0 ft =  v0 · t · sin 12° - 1/2 · 32.2 ft/s² · t²

0 ft = 200 ft/(t · cos 12°) · t · sin 12° - 16.1 ft/s² · t²       (sin 12°/cos 12° = tan 12°)

0 ft = 200 ft · tan 12° - 16.1 ft/s² · t²

(0 ft - 200 ft · tan 12°) / -16.1 ft/s² = t²

t = 1.6 s

Now, let´s calculate v0:

200 ft/(t · cos 12°) = v0

200 ft/(1.6 s · cos 12°) = v0

v0 = 127.8 ft/s

step-by-step explanation:

this is not possible to solve with the info given.   you need at least an one angle somewhere.