Question

A Stone Is Dropped From The Top Of A 40m High Tower. Calculate Its Speed After 2 Sec. Also Find The Speed With Which The Stone Strikes The Ground.

Answer 1

that's the answer hope it helps you

Answer 2

heya friend...
here is ur answer....


acc. to second law of motion
= h = ut +1/2gt^2
we have h = 40
t = 2 seconds
g = 10
u = ?
putting valies we get
40= 2u + 1/2*10*2*2
40-20 =2u
20/2 = u
speed of stone after 2 seconds = 10m/ls



Speed With Which The Stone Strikes The Ground:-
We should now examine the forces on the object by returning to our picture and isolating our object to examine forces on it.


With the object isolated by the circle, we should now use it to draw a Free-Body Diagram.


Here we see that the only force acting on the object is gravity, which means that we can solve for the acceleration of the object using Newton's Second Law.

FNet=maFNet=ma

FGravity,onStonebyEarth=mgFGravity,onStonebyEarth=mg

mg=mamg=ma

a=ga=g

And the acceleration is towards the earth, so we will at that it is negative.

a=−ga=−g, where g is the acceleration due to gravity =9.8ms2=9.8ms2

We can now find kinematics equations to model the situation, as the acceleration is constant, and use this to solve for the time, t and speed v final.

To find the final speed, we will use the equation

v2f=v2i+2aΔsvf2=vi2+2aΔs

As the object is dropped, v initial is zero, so we can solve for v final and find that

vf=±(2)(−9.8)(−40)−−−−−−−−−−−−√=−28ms=−30msvf=±(2)(−9.8)(−40)=−28ms=−30ms

HOPE IT HELPS..
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