Answer:
Explanation:
Given that,
The rising velocity of the balloon
U = 10 ft/s
The height where a small balloon is release is
y = 960ft
Taking g = - 32 ft/s²
A. Time the balloon will strike the ground.
Using the equation of free fall
∆y = ut + ½gt².
0—960 = 10t + ½× -32 × t²
- 960 = 10t —16t²
16t² — 10t — 960 = 0
Divide through by 2
8t²—5t —480= 0
Using quadratic formula method
t = [-b ± √(b²-4ac)] / 2a
a = 8 b = -5 and c = -480
t = [--5 ± √((-5)²-4×8×-480)] / 2×8
t = [5 ± √(25+15360)] / 16
t = [5 ± 124.04] / 16
Let discard the negative time since time can't be negate
Then, t = [5+124.06] / 16
t = 8.065 seconds
To 2d.p
t = 8.06 seconds
B. Final velocity it hits the ground
Using equation of motion.
V = U + gt
V = 10 + (-32) × 8.06
V = 10 — 258.07
V = —248.07m/s
The negative sign is showing that it is moving downward
The final velocity of the hot air balloon before it hits the ground is 248.1 ft/s.
The given parameters;
The time taken for the ball to strike the ground is calculated as follows;
[tex]h = ut + \frac{1}{2} gt^2\\\\960 = 10t + (0.5 \times 32)t^2\\\\960 = 10t + 16t^2\\\\16t^2 + 10t - 960 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method\\\\a = 16, \ b = 10 \ c = -960\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-10 \ \ +/- \ \ \sqrt{10^2 - 4(16\times 960)} }{2(16)} \\\\t = 7.44 \ s[/tex]
The final velocity of the hot air balloon is calculated as follows;
v = u + gt
v = 10 + (7.44 x 32)
v = 248.1 ft/s
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