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Novosadov [1.4K]
3 years ago
5

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the wa

y down and passes a point a distance 34.0 m below its starting point at a time 5.00 s after it leaves the thrower's hand. Air resistance may be ignored.
A) What is the initial speed of the egg?
B) How high does it rise above its starting point?
C) What is the magnitude of its velocity at the highest point?
E) What are the magnitude and direction of its acceleration at the highest point?
Physics
1 answer:
Lyrx [107]3 years ago
7 0

Answer:

A) 17.7 m/s

B) 15.98 m

C) Zero

E) 9.8 m/s²

Explanation:

given information

distance, h = - 34 m

time, t = 5 s

A) What is the initial speed of the egg?

h - h₀ = v₀t - \frac{1}{2} gt², h₀ = 0

- 34 = v₀ 5 - \frac{1}{2} 9.8 5²

- 34 = 5 v₀ - 122.5

v₀ = 122.5 - 34/5

    = 17.7 m/s

B) How high does it rise above its starting point?

v² = v₀² - 2gh

v = 0 (highest point)

2gh = v₀²

h = v₀²/2g

  = 17.7²/2 (9.8)

  = 15.98 m

C) What is the magnitude of its velocity at the highest point?

v = 0 (at highest point)

E) What are the magnitude and direction of its acceleration at the highest point?

g= 9.8 m/s², since the egg is moved vertically, the acceleration is the same as the gravitational acceleration.

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air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

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so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

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Area = \frac{\pi }{4} d^{2}

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force = mass × acceleration

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