Question

Two bodies are thrown
vertically upward, with the same initial velocity of 98 m/s but 4 sec apart. how
long after the first one is thrown when they meet?
(1) 10 sec
(2) I1sec
(3) 12 sec
(4) 13 sec​

Answer 1

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

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

• Initial velocity (u) = 98 m/s.
• g = - 9.8 m/s².
• Time gap between throw is = 4 seconds.

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

Let the first body be released at time "t" seconds.

and, the second body be released at time "t - 4" seconds.

Distance travelled by the first body,

Applying second kinematic equation,

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

Substituting the values,

$\large{S = ut - \dfrac{1}{2}gt^2 \: ----(1)}$

Distance travelled by the second body,

Applying second kinematic equation,

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

Substituting the values,

$\large{S = u(t - 4) - \dfrac{1}{2}g(t - 4)^2 \: ----(2)}$

When they will meet there displacement will be equal.

Now,

$\large{Equation (1) = Equation (2)}$

$\large{ut - \dfrac{1}{2} gt^2 = u(t - 4) - \dfrac{1}{2}g(t - 4)^2}$

Therefore,

$\large{ut - \dfrac{1}{2}gt^2 = ut - 4u - \dfrac{1}{2} g[t^2 + 16 - 8t}$

Now,

$\large{ut - \dfrac{1}{2}gt^2 = ut - 4u - \dfrac{1}{2} gt^2 +- \dfrac{1}{2} g(16) + \dfrac{1}{2} g(8t)}$

Therefore,

$\large{ \cancel{ut} \cancel{- \dfrac{1}{2}gt^2} = \cancel{ut} - 4u \cancel{ - \dfrac{1}{2} gt^2} - \dfrac{1}{2} g(16) + \dfrac{1}{2} g(8t)}$

It becomes,

$\large{0 = \left[ - 4u - \dfrac{1}{2}g(16) \right]+ \dfrac{1}{2}(8t)g}$

Taking L.C.M

$\large{0 = \left[ \dfrac{- 8u - 16g}{2} \right] + \dfrac{1}{2}(8t)g}$

Now,

$\large{0 = \left[ \dfrac{- 8u - 16g + (8t)g}{2} \right]}$

Now,

$\large{0 = - 8u - 16g + (8t)g}$

$\large{8u + 16g = (8t)g}$

$\large{t = \dfrac{8u + 16g}{8g}}$

Take out "8" as common factor.

$\large{t = 8 \times \left[\dfrac{u + 2g}{8g} \right]}$

$\large{t = \cancel{8} \times \dfrac{u + 2g}{ \cancel{8}g}}$

It becomes,

$\large{\boxed{t = \dfrac{u + 2g}{g}}}$

Substituting the values,

$\large{t = \dfrac{98 + 2 \times 9.8}{9.8}}$

$\large{t = \dfrac{98 + 19.6}{9.8}}$

$\large{t = \dfrac{117.6}{9.8}}$

$\large{t = \dfrac{ \cancel{117.6}}{ \cancel{9.8}}}$

It comes as,

$\huge{\boxed{\boxed{t = 12 \: Seconds}}}$

So, the time after they meet is 12 seconds (Option - (3)).