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Lady bird [3.3K]

3 years ago

13

A hot air balloon is rising vertically with a velocity of 10.0 feet per second. A very small ball is released from the hot air b

alloon at the instant when it is 960 feet above the ground. Use a(t)=−32 ft/sec2 as the acceleration due to gravity.
1. How many seconds after its release will the ball strike the ground? Round the answer to the nearest two decimal places.

2. At what velocity will it hit the ground?

1 answer:

krek1111 [17]3 years ago

7 0

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

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