wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A ball is dropped from the top of a building. The ball takes 0.5 s to fall past the length of a window, the top of the window being at a distance of 3 m from the top of the building. If the speed of the ball at the top and the bottom of the window are VT and VB respectively, then (g=9.8m/sec2)

A
VT+VB=20ms1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
VTVB=4.9ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
VBVT=1ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
VBVT=1ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A VT+VB=20ms1
From the equations of motion we know that vt2=u2+2as.
here initial velocity"u"=0 so vt=2×9.81×3=7.67m/sec
also,vb=u+at=0+9.81×0.5=4.905m/sec.
so net velocity at the bottom of window will be
VT+VB=7.67+4.905=12.57m/sec

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Equations of Motion
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon