Question

A particle is projected vertically up with speed of 10m/s . The time after which it will pass through the point above 3.2 m above the point of projection(g = 10 m/ s^2). A)0.4s
B)0.5s
C)1.2s
D)1.8s

Answer 1

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

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

u = 10 m/s.

h = 3.2 m.

g = - 10 m/s².

t = ?.

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

By applying second kinematic equation.

$\large{\bold{S = ut + \frac{1}{2}at^2}}$

Substituting the values.

$\large{ \implies S = 10t + \frac{1}{2} \times -10 \times t^2}$

$\large{ \implies 3.2 = 10t + \frac{1}{ \cancel{2}} \times - \cancel{10} \times t^2}$

$\large{ \implies 5t^2 - 10t + 3.2 = 0}$

By splitting the middle term.

$\large{ \implies 5t^2 - 8t - 2t + 3.2 = 0}$

$\large{ \implies 5t(t - 1.6) - 2(t - 1.6) = 0}$

$\large{ \implies (t - 1.6)(5t - 2) = 0}$

$\large{ \implies t = 1.6 \: and \: t = 0.4}$

$\huge{\boxed{\boxed{t = 0.4 \: seconds}}}$

As comparing t = 1.6 seconds is the time when ball is falling down.

And the question is asking the time after projection.

So, t = 0.4 seconds.

So, body will reach 3.2 m height at t = 0.4 seconds (Option--a).

Answer 2

Answer:

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Explanation:

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