David M. answered 07/16/18
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Dave "The Math Whiz"
1. The maximum height the ball will reach is the vertex of the parabola s(t)=48+80t-16t2. first, we need to solve for t. The vertex of any parabola of the form f(x)=ax2+bx+c is found by using x=-b/2a. In this case, x=t, b=80 and a=-16. Substituting we get:
t=-80/((2)(-16)--->t=-80/-32--->t=5/2
Putting this value of t in the original formula will give us the maximum height:
s(t)=48+80t-16t2
s(5/2)=48+80(5/2)-16(5/2)2
s(5/2)=48+200-16(25/4)
s(5/2)=248-100
smax=148 feet
2. The instantaneous velocity is the first derivative of the position formula (i.e. s(t)): s'(t)= -32t+80.
the time, t, is when it hits the ground, so we must solve for t when s(t)=-16t2+80t=48. Using the quadratic formula we get:
(-b±(sq root of b2-4ac))/2a
(-80±(sq root of 802-(4)(-16)(48))/((2)(-16))
(-80±(sq root of 6400+3072))/-32
(-80±(sq root of 9472))/-32
(-80±(97.32))/-32
t=-177.32/-32=5.54 and t=17.32/-32=-.54
Because time has to be positive, t=5.54 sec. Substituting this value into our first derivative equation, we get:
s'(t)=-32t+80
s'(5.54)=-32(5.54)+80
s'(5.54)=-177.28+80
s'(5.54)=-97.28 ft/sec
The answer here is -97.28 ft/sec. Why is the answer negative?
Duneshke B.
For the last part, the answer is negative because of the ball being dropped, therefore, we have a negative velocity.05/03/19