Question

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An object thrown vertically upward reaches a height of 320m. What was it's initial velocity? How long will the object take to come back to the earth? Assume g=10m/s^2
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No Spam. ​

Answer 1

Answer:

80m/s and 16s

Explanation:

$\sf{time \:of\: flight=\frac{2u}{g}}$

$\sf{u=initial\: velocity}$

$\sf{u=\sqrt{2ghmax}}$

$\sf{u=\sqrt{2×10×320}=\sqrt{6400}=80m/s}$

u=80m/s

____________________________

put value of u in formula of time of fight

$\sf{time \:of\: flight=\frac{2(80)}{10}=16s}$

time of flight=16s

Answer 2

Answer :-

→ Initial velocity of the object is 80 m/s .

→ It takes 16 sec to come back to earth .

Explanation :-

We have :-

• Maximum height reached (h) = 320 m

• Gravitational acceleration (g) = 10 m/s²

______________________________

Firstly, when we consider the journey of the object vertically upwards :-

• Acceleration (g) will be -10 m/s²

• Final velocity (v) will be 0 m/s .

Initial velocity :-

Putting values in the 3rd equation of motion :-

⇒ v² - u² = 2gh

⇒ 0 - u² = 2(-10)(320)

⇒ -u² = -6400

⇒ u = √6400

⇒ u = 80 m/s

Time of ascent :-

By using the 1st equation of motion :-

⇒ v = u + at

⇒ 0 = 80 + (-10)t

⇒ -80 = -10t

⇒ t = 8s

As we know that Time of ascent = Time of descent , so the object will take 8 seconds to descend from a height of 320 metres .

Time taken to come back to Earth :-

= Time of ascent + Time of descent

= (8 + 8) s

= 16 s