Problems 2.114-115 Example 2-21 depicts the following scenario with the accompan

Question

Problems 2.114-115

Example 2-21 depicts the following scenario with the accompanying figures. A hot-air balloon is rising straight upward with a constant speed of 6.5 m/s. When the basket of the balloon is 20.0 m above the ground, a bag of sand tied to the basket comes loose and reaches a maximum height of 22 m.

Now we will consider some slightly different related scenarios to Example 2-21.

Part A

If the bag in Example 2-21 had been released at 31.0 m instead, with everything else remaining the same, would its maximum height be 33 m, greater than 33 m,or less than 33 m?

If the bag in Example 2-21 had been released at 31.0  instead, with everything else remaining the same, would its maximum height be 33 , greater than 33 , or less than 33 ?

SubmitMy AnswersGive Up

Part B

Find the speed of the bag in part A just before it lands when it is released from 31.0 m.

SubmitMy AnswersGive Up

Part C

Now, suppose the balloon in Example 2-21 is descending with a constant speed of 4.2 m/s when the bag of sand comes loose at a height of 35 m. What is the amount of time the bag is in the air?

Express your answer using two significant figures.

SubmitMy AnswersGive Up

Part D

What is the speed of the bag when it is 16 m above the ground?

Express your answer using two significant figures.

Explanation / Answer

Given,

v = 6.5 m/s ; y0 = 20 m ; y = 22 m

A)since the ballon is moving up with the constant speed,

h = v^2/2g = 6.5^2/2 x 9.8 = 2.2 m

H = 2.2 + 31 = 31.2 m

for this we can say, greater than equal to 33 m

B)from eqn of motion

v^2 = u^2 + 2 a s

as it touches the ground v becomes zero, but u the speed just before it hits:

u = sqrt (2 a s) = sqrt (2 x 9.8 x 31) = 24.65 m/s

Hence, v = 24.65 m/s

C)from eqn of motion

S = ut + 1/2 at^2

35 = 4.2 t + 0.5 x 9.8 x t^2

4.9 t^2 + 4.2 t - 35 = 0

the abive quadratic equation gives us:

t = 2.28 s and -3.14 sec ; ignoring negative value

Hence, t = 2.28 s

D)from v^2 = u^2 + 2 a s

v^2 = 4.2^2 + 2 x 9.8 x 16 = 331.24

v = 18.2 m/s

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.