Question

An object is thrown vertically upwards from the ground at 30m/s.what is the maximum height it attains?

Answer 1

Answer:

Maximum height attained by the object is 45m.

Explanation:

From : Equations of motion -

• v^2 = u^2 + 2aS, where S is the total displacement, u is the initial velocity, a is the acceleration in the body and v is the final velocity.

Here,

Object, which was at rest, is now thrown vertically upwards with velocity 30 m / s. Since it is moving in vertically upward direction, acceleration is now gravitational acceleration( - g )opposite to the astronomical body, earth.

Thus, here, a = - g.

At maximum height, velocity of the body = 0.

Thus,

= > v^2 = u^2 + 2aS

= > 0^2 = ( 30 m / s )^2 - 2gS

= > 0 = 900 m^2 / s^2 - 2( 10 m / s^2 )S { g is taken as 10 m/s^2 }

= > 900 m^2 / s^2 = 20 m / s^2 x S

= > 900 / 20 m = S

= > 45 m = S

Hence the maximum height attained by the object is 45m.

Answer 2

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The maximum height attained by the object is 45.91 metres.

$\huge{\purple{\underline{\underline{\mathfrak{♡Êxplanation♡}}}}}$

Given :

• u = 30 m/s
• v = 0
• g = -9.8 m/s^2

To find :

• s = ?

Now,

${v}^{2} - {u}^{2} = 2gs \\ = > {(0)}^{2} - {(30)}^{2} = 2 \times ( - 9.8) \times s \\ = > - 900 = - 19.6 \times s \\ = > s = \frac{ - 900}{ - 19.6} \\ = > s = 45.91 \: m$

Hence, the object will attain a height of 45.91 metres.

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