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

A firecracker is thrown with velocity of 30 m.s1 in a direction which makes an angle of 75o with the vertical axis. At some point on its trajectory, the firecracker splits into two identical pieces in such a way that one piece falls 27 m far from the shooting point. Assuming that all trajectories are contained in the same place, how far will the other piece fall from the shooting point ? (Take g = 10 m.s2 and neglect air resistance)

A
63 m or 144 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
28 m or 72 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
72 m or 99 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
63 m or 117 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is C 63 m or 117 m
θ=15
Range=v2sin2θg=45m
Since there is no horizontal force acting so the final CM should remain at 45m.
Let say the particle landed 27m away from initial point toward CM
Then m×27+m×x2m=45x=63m
if the object landed on other side of origin
m×27+m×x2m=45x=117m

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon