Question

an object is thrown up with speed 49 m/s. At what instants is it 19.6 m below the higest point..? [ take g = 9.8m/s²]​

Answer 1

GIVEN:-

• $\rm{Initial\:Velocity = 49m/s}$

• $\rm{Acceleration = -9.8m/s^{-2}}$

• $\rm{Final\:Velocity = 0m/s}$.

TO FIND:-

The time when object is 19.6m below the highest point it reached.

FORMULAE USED:-

• ${\boxed{\rm{ v^2 - u^2 = 2as}}}$

• ${\boxed{\rm{ S = ut+\dfrac{1}{2}\times{a}\times{(t)}^2}}}$.

Now,

Usiing the Third equation of motion to find the maximum Height.

$\implies\sf{v^2 - u^2 = 2as}$

$\implies\rm{ (0)^2 - (49)^2 = 2\times{-9.8}\times{S}}$

$\implies\rm{ -2401 = -19.6s}$

$\implies\rm{ S = \dfrac{-2401}{19.6}}$

$\implies\rm{ S = 122.5m}$

According to question,

• The height of the object when it is 19.6m below its highest Point - 122.5 - 19.6m = 102.9m.

• Now, We will use Second equation of motion to find the time taken by object to reach there.

Now,

$\implies\rm{S = ut+\dfrac{1}{2}\times{a}\times{(t)}^2}$

$\implies\rm{ 102.9m = 49\times{t} + \dfrac{1}{2}\times{-9.8}\times{(t)^2}}$

$\implies\rm{ 102.9m = 49t - 4.9t^2 }$

$\implies\rm{ 102.9 = 4.9t (10 - t)}$

$\implies\rm{\dfrac{102.9}{4.9} = 10t - t^2}$

$\implies\rm{ 21- 10t + t^2}$

$\implies\rm{ t^2-10t +21}$

$\implies\rm{ t^2 -7t-3t +21}$

$\implies\rm{ t(t-7) -3(t-7)}$

$\implies\rm{ t -3 = 0}$

$\implies\rm{ t = 3s}$

Hence, The object is 19.6m below from the highest Point in 3rd second