A robot probe drops camera off the rim of a 532 m high cliff

A robot probe drops camera off the rim of a 532 m high cliff, where the free-fall acceleration is -3.7 m/s^2. Find velocity which it hits the ground.

answer in units of m/s.

Solution

(With negative free fall acceleration , the camara cannot move towards the ground. A correction is done at the end .The equation of motion does not lead us wrong.)

The equation of motion of the free falling camera is given by

v^2-u^2=2as. where s is the vertical displacement due to the free fall acceeration and u is the initial velocity and v is the final velocity while reaching the ground and a is the given free fall aceleration .

Here u=0 , the initial velcity , a= -3.7m/s^2 and s=538

V^2-0 = 2(-3.7) 532

v = square root of (2*-3.7*532)=sqrt(-3936.8) which is imaginary.

Under a negative acceleration , the camara flies away. It does not reach the ground.

So the acceleration of the free fall must be with a positive acceleration of 3.7 m/s^2. with this correction , if you revist the problem, we get v=sqrt(3936.8) = 62.74 m/s while reaching the ground.