Question

An object is thrown vertically from a height of $ 15m $ at $ 6\dfrac{m}{s} $ . How long will it take for the object to hit the ground?

Answer 1

Hint :We are given a height from which the object is thrown vertically upwards . First we will calculate the time to get the object back to its original height then we will calculate the time taken to get to the ground. Remember that when the object comes back to it height it will have the same velocity it had when it went up, We are only considering the normal vertical motion and therefore we will not consider the horizontal motion equation. The time will therefore be calculated in two parts ,
 $ {t_1} $ is the time taken to reach the same height,
And $ {t_2} $ will be the time taken to reach ground ,
 $ {t_1}and{\text{ }}{t_2} $ are given by following equation which are form the equation of vertical motions,
 $ {t_1} = \dfrac{{2u}}{g} $
And $ {t_2} = \dfrac{{v - u}}{g} $ the final velocity $ v $ will be calculated by
 $ v = \sqrt {{u^2} + 2aS)} $
Where $ s $ is the distance.

Complete Step By Step Answer:
First we will have to calculate the value $ {t_1} $ which is the time taken to reach the same height,
We will take $ g = 9.8m{\sec ^{ - 2}} $
 $ {t_1} = \dfrac{{2u}}{g} $
 $ \dfrac{{2u}}{g} = \dfrac{{2 \times 6m{s^{ - 1}}}}{{9.8m{s^{ - 2}}}} $
 $ \dfrac{{2 \times 6m{s^{ - 1}}}}{{9.8m{s^{ - 2}}}} = 1.2s $
Thus the value of $ {t_1} = 1.2s $
Now we will calculate the value of $ {t_2} $ which will be the time taken to reach ground ,
First we will calculate the final velocity,
 $ v = \sqrt {{u^2} + 2aS} $
 $ v = \sqrt {{{\left( {6m{s^{ - 1}}} \right)}^2} + \left( {2 \times 9.8m{s^{ - 2}} \times 15m} \right)} $
Solving the question further,
 $ v = 18.2m{s^{ - 1}} $
Now calculating $ {t_2} $
 $ {t_2} = \dfrac{{v - u}}{g} $
 $ \dfrac{{v - u}}{g} = \dfrac{{18.2m{s^{ - 1}} - 6m{s^{ - 1}}}}{{9.8m{s^{ - 2}}}} $
Solving the equation we get,
 $ \dfrac{{18.2m{s^{ - 1}} - 6m{s^{ - 1}}}}{{9.8m{s^{ - 2}}}} = 1.2s $
 $ {t_2} = 1.2s $
Thus total time taken will be given by,
 $ {t_1} + {t_2} = 1.2s + 1.2s $
 $ {t_1} + {t_2} = 2.4s $
Which is our final answer.

Note :
If a initial velocity is given, the acceleration is uniform and displacement is given and we have to get the final velocity we will use the formula,
 $ v = \sqrt {{u^2} + 2aS} $