Question

a ball is thrown upwards from the ground with a velocity of 30 m/s . How long does it take ti reach its highest point ? what is its height above GROUND? ​

Answer 1

Given, u = 30m/s

We know, v = 0 m/s and a = 9.8 m/s^2

Then, using Newton's formula;

v = u + at

or, 0 m/s = (30 + 9.8t) m/s

or, t = -30/9.8 s

or, t = 3.06 s (approx.)

Now, again using the other formula;

s = ut + 1/2 at^2

or, h = 94.80 + 1/2 • 9.8 • 9.3636

or, h = 94.8 + 45.88164

or, h = 140.68 (approx.)

HOPE THIS COULD HELP!!!

Answer 2

$\huge{\bold{\underline{\underline{ ....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Initial velocity (u) = 30 m/s.
• Final velocity (v) = 0 m/s.
• Acceleration due to gravity = - 10 m/s².

$\huge{\bold{\underline{Explanation:-}}}$

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From First Kinematics equation.

$\large{\boxed{\tt v = u + at}}$

Substituting the values,

$\large{\tt 0 = 30 + (- 10) \times t}$

$\large{\tt 0 = 30 - 10t}$

$\large{\tt 10t = 30}$

$\large{\tt t = \dfrac{30}{10}}$

$\large{\tt t = \dfrac{\cancel{30}}{\cancel{10}}}$

$\large{\tt t = 3 \: Seconds \: ----(1)}$

$\huge{\boxed{\boxed{\tt t = 3 \: seconds}}}$

So, the time taken to reach the highest point is 3 seconds.

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Applying second kinematic equation,

$\large{\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

Substituting the values,

And substituting the value of time from equation (1).

$\large{\tt S = 30 \times 3 + \dfrac{1}{2} \times - 10 \times (3)^2}$

Now,

$\large{\tt S = 30 \times 3 - \dfrac{1}{2} \times 10 \times 9}$

$\large{\tt S = 90 - \dfrac{10 \times 9}{2}}$

$\large{\tt S = 90 - \dfrac{90}{2}}$

$\large{\tt S = 90 - \dfrac{\cancel{90}}{\cancel{2}}}$

$\large{\tt S = 90 - 45}$

$\huge{\boxed{\boxed{\tt S = 45 \: meters}}}$

So, the Height attained by the body is 45 meters.

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Note:-

• Here we have taken final velocity as zero because at the topmost poin its velocity is zero.
• And, acceleration is taken as negative because, it is moving against the gravitational pull(Force).

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