a certain freely falling object requires 1.05 s to travel the last 33.0 m before it hits the ground. from what height above the ground did it fall?

Using the concept of motion, we find that  38.40m is the height of cliff from where object is dropped.Let suppose total height above the ground did it fall is x mSuppose the point is P when object is to travel 33.mWe know very well that when ball hit the ground its final velocity will be zero.So, we apply the first law of motion at point to calculate the initial velocity of objectv=u + at------------(eq1)where, v is final velocity and u is initial velocity and a is accelerationputting values  in eq1we know that g=9.8m[tex]s^{-2}[/tex]=>0=u-9.8×1.05  =>u= 9.8×1.05=>u=10.29m[tex]s^{-2}[/tex]Now, we apply third law of motion to point P from the top point of cliff.Here initial velocity will become final velocity at point P.Therefore,[tex]v^{2}-u^{2}=2 a S[/tex] where, v is final velocity and u is initial velocity and a is accelerationInitial velocity(u)=0Therefore,[tex]v^{2}-0=2.9.8.S[/tex]         =>(10.29)×10.29=2×9.8×S         =>S=(105.8841)/19.6          =>S=5.40mTotal height is=S+33=38.40mHence, the height of cliff from where object is dropped =38.40mTo know more about motion, visit here:https://brainly.com/question/2281047#SPJ4;