CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A body dropped from the top of a tower falls through 40 m during the last two seconds of its fall. The height of tower is (g=10 m/s2)

A
60 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
45 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
80 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
50 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 45 m
Let the body fall through the height of tower in t seconds.
We know that distance covered the body (Dn) at a particular second n is given by the relation
Dn=u+a2(2n1)
As the body is at rest initially
u=0
Dn=a2(2n1)
Let t be the time taken by the body to reach ground.
Distance travelled in last 2 seconds =40 m (given)
Total distance travelled in last 2 seconds of fall can also be expressed as
D=Dt+D(t1)
=[0+g2(2t1)]+[0+g2{2(t1)1}]
=g2(2t1)+g2(2t3)=g2(4t4)
=102×4(t1)
40=20(t1)
t=2+1=3 s
Distance travelled in t seconds is
s=ut+12at2=0+12×10×32=45 m

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Motion Under Gravity
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon