Question

an object is thrown vertically upwards and rises to hight of 13.07 calculate:
I) the velocity with which the object was thrown upwards
ii) the time taken by the object to reach the highest point ​

Answer 1

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＊ An object is thrown vertically upwards and rises to hight of 13.07m

Calculate:

i) The velocity with which the object was thrown upwards.

ii) The time taken by the object to reach the highest point.

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‣ Height, s = 13.07m

‣ Final Velocity, v = 0m/s

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˃ Initial velocity, u

˃ Total time taken to reach the highest point

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(i)

Here,

we, know that

$\underline{\boxed{\sf v^2 - u^2 = 2as}} \qquad \quad \big\lgroup { \sf 2^{nd} \: Equation \: of \: Motion } \big\rgroup$

where,

• Final Velocity, v = 0m/s
• Acceleration = -g = -9.8m/s²
• Distance, s = 13.07m

So,

⇰v² - u² = 2(-g)s

⇰0² - u² = 2(-9.8)(13.07)

⇰- u² = -19.6(13.07)

⇰-u² = −256.172

⇰u² = 256.172

⇰u = $\sqrt{256.172}$

⇰u = 16.0053(aaprox.)m/s

⇰u = 16m/s $\red\bigstar$

Hence, velocity at which the object was thrown is 16m/s

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(ii)

Here, we know that

$\underline{\boxed{\sf v = u+ at}} \qquad \quad \big\lgroup { \sf 3^{rd} \: Equation \: of \: Motion } \big\rgroup$

where,

• Final Velocity, v = 0m/s
• Initial Velocity, u = 16m/s
• Acceleration, a = -g = -9.8m/s²

So,

⤚v = u + (-g)t

⤚0 = 16 + (-9.8)t

⤚-16 = -9.8t

⤚16 = 9.8t

⤚$\dfrac{16}{9.8}$ = t

⤚1.632(aaprox.)sec = t

⤚1.6sec = t

⤚t = 1.6sec $\pink\bigstar$

Hence, time taken by object to reach the maximum height is 1.6sec

Answer 2

Answer:

this is picture will help you

Explanation:

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