Question

A ball is thrown straight upward at an initial speed of
v0 = 80 ft/s.
(Use the formula
h = −16t2 + v0t.
If not possible, enter IMPOSSIBLE.)
(a) When does the ball initially reach a height of 64 ft?
1

Correct: Your answer is correct.
s

(b) When does it reach a height of 128 ft?
s

(c) What is the greatest height reached by the ball?
ft

(d) When does the ball reach the highest point of its path?
s

(e) When does the ball hit the ground?
s

Answer 1

h = −16t^2 + v0*t, where v0 is given as 80 ft/s

so your equation becomes h = -16t^2 + 80t

easiest way to solve is by graphing. however, it is also good to know how to do these by hand.

I assume you got a) correct, so I'll move onto b)

b) set h = 128 and solve. use the quadratic formula (in this case, your c-value is 0 as there is no constant term). as a major hint, think about what it means for a quadratic equation when the discriminant is negative.

I'll actually solve d) before c), as d) will help us with c).

d) you're looking for the highest point, or the vertex. the x-coordinate of the vertex is -b/(2a). plug your a and b values into this and evaluate.

c) take the time from d), and plug it into the height equation

e) at the ground, h = 0. set h = 0 and solve for t. you'll have two solutions, one of which is the trivial solution t = 0, but you want the other one.

Answer 2

1.
\[64=-16t^2+80t\]
\[4=-t^2+5t\]
\[t^2-5t+4=0\]
\[t^2-4t-t+4=0\]
t(t-4)-1(t-4)=0
(t-4)(t-1)=0
t=4,1
after 1 s and 4 s it is at height of 64 ft.