A stone is dropped from the top of14m high tower calculate its speed after 2 seconds . also find the speed with which the stone strikes the ground.
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Answered by
1
Hello mate this is the solution,
we have,
distance,s=14m
initial velocity,u=0 ,because it gently dropped,
final velocity=v
time,t=2 seconds
acceleration due to gravity,g=9.8ms^-2
so ,by using first equation of motion:
v=u+at
v=0+9.8*2
v=19.6ms^-1
now ,using second equation of motion:
v=1/2at^2
v=1/2*9.8*2^2
v=1/2*9.8*4
v=19.6m/s
the velocity just before hitting..
hope this answer helps you..
thanks for questioning
we have,
distance,s=14m
initial velocity,u=0 ,because it gently dropped,
final velocity=v
time,t=2 seconds
acceleration due to gravity,g=9.8ms^-2
so ,by using first equation of motion:
v=u+at
v=0+9.8*2
v=19.6ms^-1
now ,using second equation of motion:
v=1/2at^2
v=1/2*9.8*2^2
v=1/2*9.8*4
v=19.6m/s
the velocity just before hitting..
hope this answer helps you..
thanks for questioning
yadavakilesh2074:
hey bro can you plzz explain me how much time will it take for its entire motion??if we use equation s=ut+1/2at^2 t=1.67 whcih is wrong plzz let me know how to do it?
brother but time is given 2 seconds,why we use 1.67 seconds.
brother if we assume that time is not given and we try to find out it by using kinematical formula then it comes 1.67 so where I am wrong???I thought initially they have ask v for first 2 sec but what will be the time for entire journey??
Answered by
1
v=u+at
so,v=g*2=9.8*2=19.6
speed with which it strikes the ground..
s=ut+1/2at²
14=1/2gt²
so t=1.69sec
so by using v=u+at we get v=16.56 you can verify it from the other kinematic formula v²=u²+2as
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