Question

A ball is dropped from the top of a 100m high tower on a planet. In the last $ \dfrac{1}{2} $ s before hitting the ground, it covers a distance of 19m. Acceleration due to gravity (in $ {\text{m/}}{{\text{s}}^2} $ ) near the surface on the planet is:______

Answer 1

Hint: Since in the final half second, the stone travelled 19 m, then the stone travelled 81 m before then. The final velocity at the 81 m mark, is the initial velocity for the last 19 m (which took a time of half seconds).

Formula used: In this solution we will be using the following formulae;
 $ {v^2} = {u^2} + 2gs $ where $ v $ is the final velocity of a stone travelling vertically downward, $ u $ is the initial velocity, $ g $ is the acceleration due to gravity of the planet, and $ s $ is the distance travelled.
 $ s = ut + \dfrac{1}{2}g{t^2} $ where $ s $ remains the distance travelled, and $ t $ is time taken to cover that distance.
 $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ for a quadratic equation $ a{x^2} + bx + c = 0 $

Complete Answer:
In the question, we are told that the ball travelled a distance of 19 m for the last half of a second, hence, before then, the ball has travelled
 $ 100 - 19 = 81m $
The final velocity, when the 81 m journey ended, is the initial velocity of the last 19 m.
Now, from equations of motion, we have that
 $ {v^2} = {u^2} + 2gs $ where $ v $ is the final velocity of a stone travelling vertically downward, $ u $ is the initial velocity, $ g $ is the acceleration due to gravity of the planet, and $ s $ is the distance travelled.
Hence,
 $ {v^2} = 0 + 2g\left( {81} \right) $
 $ \Rightarrow v = \sqrt {2 \times 81g} = 9\sqrt {2g} {\text{ m/s}} $
Now, for the final 19 m, it can be defined using the equation
 $ s = ut + \dfrac{1}{2}g{t^2} $ where $ s $ the distance travelled, $ u $ remains the initial velocity, and $ t $ is time taken to cover that distance.
Hence,
 $ 19 = 9\sqrt {2g} \left( {\dfrac{1}{2}} \right) + \dfrac{1}{2}g{\left( {\dfrac{1}{2}} \right)^2} $ (recall final velocity of 81 m journey is initial velocity of 19 m journey)
 $ \Rightarrow 19 = \dfrac{9}{2}\sqrt 2 \sqrt g + \dfrac{g}{8} $
Let $ f = \sqrt g $ , then
 $ \Rightarrow 19 = \dfrac{9}{2}\sqrt 2 f + \dfrac{{{f^2}}}{8} $
Multiply through by 8, we have
 $ 152 = 36\sqrt 2 f + {f^2} $
 $ \Rightarrow {f^2} + 36\sqrt 2 f - 152 = 0 $
Using, the quadratic formula, given by
 $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ for an equation $ a{x^2} + bx + c = 0 $
we have
 $ f = \dfrac{{ - 36\sqrt 2 \pm \sqrt {{{\left( {36\sqrt 2 } \right)}^2} - 4\left( 1 \right)\left( { - 521} \right)} }}{2} $
Calculating and simplifying we would have
 $ f = 2\sqrt 2 $ (we have neglected the negative answer as it makes no sense, since acceleration due to gravity should be positive)
Now, since $ g = {f^2} $ , we have
 $ g = {\left( {2\sqrt 2 } \right)^2} = 8m/{s^2} $ .

Note:
For clarity, a negative acceleration due to gravity signifies that it points upward. This is because we have assumed positive is downward when we defined our equations of motion as
 $ {v^2} = {u^2} + 2gs $ and not $ {v^2} = {u^2} - 2gs $ , nor did we make our velocity values negative.