After how many seconds is the height 48 feet? (Hint:Let h=48 and solve for t.)?

Question

If an object is propelled upward from the ground level with an inital velocity of 64 feet per second, its height h in feet, t seconds later is given the equation h=-6t2+64.

Answer 1

h = 6t² + 64

Set h = 48
48 = -6t² + 64

Subtract 64 from both sides:
-16 = -6t²

Divide both sides by -6:
16/6 = t²
8/3 = t²

Take the square root of both sides:
t = sqrt(8/3)
t = sqrt(24/9)
t = sqrt(4)*sqrt(6) / sqrt(9)
t = 2 * sqrt(6) / 3
t = 2/3 sqrt(6)
t ≈ 1.63299316 seconds

Answer 2

The equation supplied would describe an object being dropped from a height of 64 feet, rather than propelled from the ground, but anyway, doing what they say:

Start with the equation:
h = 6t² + 64

Set h = 48
48 = -6t² + 64

Subtract 64 from both sides:
-16 = -6t²

Divide both sides by -6:
16/6 = t²
8/3 = t²

Take the square root of both sides:
t = sqrt(8/3)
t = sqrt(24/9)
t = sqrt(4)*sqrt(6) / sqrt(9)
t = 2 * sqrt(6) / 3
t = 2/3 sqrt(6)
t ≈ 1.63299316 seconds

So after approximately 1.63 seconds, the ball will have fallen from 64 feet to 48 feet.

Answer 3

Where u is the initial velocity, the formula should be
h = ut – 16t^2
For a height of 48 ft,
48 = 64t – 16t^2
Solving,
t^2 – 4t + 3 = 0
(t – 1)(t – 3) = 0
t = 1 and 3
So the height is 48 ft at
1 second and again at 3 seconds.

Answer 4

You've garbled the equation. Without gravity, its height would be h = 64t (not just 64) since it would just keep on rising. The effect of gravity adds -16t² (not -6t², unless you're on another planet). So solve

48 = -16t² + 64t
16t² - 64t + 48 = 0
t² - 4t + 3 = 0
(t - 3)(t - 1) = 0
t = 1 s. or t = 3 s.
(once on the way up, once on the way down)