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). 31.43°

B). 225 feet

Step-by-step explanation:

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

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

Then the equation will be

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 = 112.5 feet

c

step-by-step explanation:

i'm sorry mate, but i am not seeing any figures. i am unable to solve this for you.