PERFECT application of the general equation of things
that are moving up and down in gravity:
H(t) = H₀ + v₀t + 1/2 g t²
t = time after the toss
H(t) = height at any time after the toss
H₀ = height from which it's tossed
v₀ = initial speed
g = acceleration of gravity
I'm going to call 'UP' = the positive direction .
So ...
H₀ = +40 ft
v₀ = +54 ft/sec
g = -32 ft/sec²
When the object hits the ground, H(t) = 0 .
So you have ...
0 = 40 + 54 t - 16 t²
Let me re-write that for you, to make it a
little easier and more familiar:
- 16 t² + 54 t + 40 = 0
Divide each side by -2 : 8 t² - 27 t - 20 = 0
Now go to your toolbox, pull out your quadratic formula,
solve that quadratic equation, and you'll get the usual two
solutions.
The positive one is the one you want.
The negative one refers to time before the toss, and may be discarded.
Answer: 6 cm
Step-by-step explanation:
Answer:
The system has unique solution.
Using backward substitution we have
1. 
2.
3.
4.
The the solution is 
