Weird planet!

s= ut + 0.5 g t^2 where s = 0m as rock return to the hand
so g = 1.5 ms^(-2)

We know that $acceleration=\frac{final velocity - initial velocity}{time}$ Now, by plugging in, we get: $a=\frac{0-15 m/s+(-15 m/s-0)}{20.0 s}$ Therefore, $a=(-30 m/s)/20 s$ or $a=\boxed{1.5 m/s^2}$ (That's positive because we are dealing with acceleration, not deceleration. Also, the author of the problem said that we should neglect the negative sign, in the first place.).

time taken to come back after going up= 2u/g. thus: 2x15/g=20. thus g=30/20=1.5

Given : u=15 m/s ; v=0; t=20 seconds. Time taken for half the journey = 20/2 =10 seconds. We know, v=u-gt ; 0=15-(10*g); Therefore, g= 1.5m/(s^2)

When the brick takes 20 sec. To go up & returns, it means that it takes 10sec. To go up only. Therefore, a = (Vf -Vi)/t = (0 - 15) / 10 = 1.5 m/s^2 (neglecting the negative sign)

we can use the first law of motion: (final velocity) = (initial velocity) + (acc.)*(time),

the rocks time to reach the max. height = 20/2 = 10 sec

the initial velocity = 15 m/s

the final velocity at the max. height = 0 m/s

so by substituting in the LAW: 0 = 15 + (Acc.)(10) -15 = (Acc.)(10) (Acc.) = -15/10 = -1.5 m/s^2

Neglecting the (-ve) sign: 1.5 m/s^2

I used the fact that if the object took 20 seconds to go up and and return, it took 10 seconds to reach its maximum height, which also means it took 10 seconds for its speed to reduce from 15m/s to 0. Acceleration= (change in velocity)/time = 15/10=1.5m/s^2

To solve this, we use this simple equation of motion: $a\quad =\quad \frac { v\quad -\quad u }{ t }$

v denotes final velocity.

u denotes initial velocity.

a denotes acceleration.

t denotes the time.

Assuming there are no frictional forces, we know that the final velocity as it touches his hand is $-15{ ms }^{ -1 }$ . Plugging in the values will give acceleration to be -1.5 ${ ms }^{ -2 }$ . Since we are neglecting negative values, the answer is $\boxed{1.5{ ms }^{ -2 }}$