Answer:
8 m sā»Ā²
Explanation:
To find the acceleration due to gravity near the surface of the planet, we first need to calculate the velocity of the ball when it is 19 m before hitting the ground.
To do this, we can use the kinematic equations of motion (SUVAT equations):
[tex]\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{array}{l}\textsf{$s$ = displacement in m}\\\\\textsf{$u$ = initial velocity in m\:s$^{-1}$}\\\\\textsf{$v$ = final velocity in m\:s$^{-1}$}\\\\\textsf{$a$ = acceleration in m\:s$^{-2}$}\\\\\textsf{$t$ = time in s (seconds)}\end{array}}[/tex]
As the ball is dropped from the top of a 100 m tower, we will take the downward motion as positive.
The ball's initial velocity (u) is 0 m sā»Ā¹ and the distance (s) it has travelled when it is 19 m above the ground is 100 - 19 = 81 m. Therefore:
Substitute these values into vĀ² = uĀ² + 2as and rearrange to isolate v:
[tex]v^2 = 0^2 + 2a(81)\\\\v^2=162a\\\\v=\sqrt{162a}[/tex]
Therefore, the velocity of the ball at the instant it is 19 m above the ground is [tex]\sqrt{162a}[/tex] m sā»Ā¹.
Now, we can use another kinematic equation for the ball's journey in the last 1/2 s before hitting the ground to find the acceleration due to gravity (a). Ā In this case:
Substitute these values into s = ut + (1/2)atĀ² and solve for a:
[tex]19 = \sqrt{162a}(0.5) + \dfrac{1}{2}a(0.5)^2\\\\\\19=\dfrac{\sqrt{162a}}{2}+\dfrac{a}{8}\\\\\\19=\dfrac{4\sqrt{162a}}{8}+\dfrac{a}{8}\\\\\\19=\dfrac{4\sqrt{162a}+a}{8}\\\\\\152=4\sqrt{162a}+a\\\\\\152-a=4\sqrt{162a}\\\\\\(152-a)^2=(4\sqrt{162a})^2\\\\\\23104-304a+a^2=2592a\\\\\\a^2-2896a+23104=0\\\\\\a=8, a=2888[/tex]
If we substitute a = 2888 back into the displacement equation we get s = 703 m, whereas if we substitute a = 8, we get s = 19 m. Therefore, the acceleration due to gravity (in m sā»Ā²) is:
[tex]\LARGE\boxed{\boxed{\rm 8 \; m\;s^{-2}}}[/tex]