wiz-icon
MyQuestionIcon
MyQuestionIcon
16
You visited us 16 times! Enjoying our articles? Unlock Full Access!
Question

An elevator without a ceiling is ascending with a constant speed of 10m/s. A boy on the elevator shoots a ball directly upward, from a height of 2.0m above te elevator floor. At this time the elevator floor is 28m above the ground. The initial speed of the ball with respect to the elevator is 20m/s. (g= 10m/s2). What maximum height above the ground does the ball reach?

A
45
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
30
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
75
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
90
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 75
absolute initial velocity of the ball is 20+10=30 m/s, thus the max height reached from the point of projection is 45 m. point of projection is itself at a height of 28+2 = 30 m. thus, max height from the ground is 45+30=75m.

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