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

A projectile is thrown with velocity u making angle θ with the vertical. It just crosses the top of two poles each of height h after 1 sec and 3 sec respectively. The maximum height of projectile is

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

The correct option is B 19.6 m

The projectile is at height h at times 1 sec and 3 sec, means the total time of projectile is 4 sec, since it takes the same 1 sec to reach ground after that 3 secs.

So it will be at Maximum height at time t = 2 sec

H=12gt2
H=12×9.8×2×2
H = 19.6 m

flag
Suggest Corrections
thumbs-up
74
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon