Jennifer hit a golf ball from the ground and it followed the projectile...?

a.
Method 1: Completing square

h(t) = -16t² + 100t
h(t) = -16(t² - 6.25t)
h(t) = -16[t² - 6.25t + (6.25/2)²] + 16×(6.25/2)²
h(t) = -16(t - 3.125)² + 156.25

When t is any real number, -16(t - 3.125)² ≤ 0
Hence, h(t) = -16(t - 3.125)² + 156.25 ≤ 156.25
The highest height = 156.25 ft ≈ 156 ft


Method 2: Differentiation

h(t) = -16t² + 100t
h'(t) = -32t + 100 = -32(t - 3.125)
h"(t) = -32

When t = 3.125:
h(t) = -16(3.125)² + 100(3.125) = 156.25
h'(t) = 0
h"(t) = -32 < 0
Hence, at t = 3.125 s, the highest height = 156.25 ft ≈ 156 ft

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b.
When h(t) = 30:
30 = -16t² + 100t
16t² - 100t + 30 = 0
8t² - 50t + 15 = 0
t = [50 ± √(50² - 4×8×15)] / (2×8)
Time taken, t = 5.93 s   or   t = 0.32 s

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c.
When h(t) = 0
0 = -16t² + 100t
t² - 6.25t = 0
t(t - 6.25) = 0
t ≠ 0, then time taken, t = 6.25 s

There are many ways to solve this: Via highschool algebra, physics/kinematic, calculus...

I will go with the first.

a) Your h(t) is a downwards parabola (why?). It therefore has a max, which is its vertex.
Whats the formula for the vertex of a parabola? Or what procedure you learned to find the max?

b) You are given that h(t) = 30. Thus -16t² + 100t = 30. How do you solve a quadratic eqs?

c) Meaning h(t) = 0. Thus -16t² + 100t = 0 . How can you solve that? Hint: factor out t. What does that give?

Answer the above and then we can continue to go through the details.


Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them.