Answer:
53
Explanation:
i used an angle of 45 degrees.
Distance from ground = 3feets
Angle = 45⁰
U is the initial velocity = 80ft/sec
g = 32
The question wants us to vfind maximum height
We have
r(t) = (80cos45⁰)ti + [3 + 80sin45⁰]t - 0.5(32)t²
We know cos 45 is also sin45 = 1/√2
r(t) = (40√2)ti + [3+40√2]t - 16t²
When we differentiate with respect to t:
r'(t) = (40√2)i + (40√2 - 32t)
= 40√2 =32t
t = 40√2/32
t= 20√2/16
We put the value of t into (3 + 40√2)t -16t²
When we substitute and simplify this we have
3 + 56.57 + 1.77 -16 +8
= 53.34
Which is approximately 53.