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 96 ft?
s

(b) When does it reach a height of 192 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?

Answer 1

@nishio @astrid1

Answer 2

(a) To find the time when the ball reaches a height of 96, use the formula given; substitute "h" for the height, 96, substitute v0 for the initial velocity, 80 ft/second, and use algebra to solve for "t", time in seconds
(b) Use the same method as for (a); keep in mind that it may be impossible for the ball to reach this height
(c) & (d) The greatest height should occur when 32t = v0; in other words, t = 88/32; I reached this result based on derivative with calculus, I can explain in more detail if you'd like; and if you plug this time into the formula given, you should be able to calculate the maximum height.
(Apologies, this is probably not the method intended for this problem, but I'm not sure exactly what you're supposed to use to solve parts c and d; the answer should be the same, regardless)
(e) Use the same methods as for (a) and (b), but replace the height with 0 in order to represent the ball reaching the ground; you could also say 16t = 88 and solve for "t"