A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of 3 ms-2 for 0.5 min. If the maximum height reached by it is 80 m, then the angle of projection is (g = 10 ms-2)
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
tan-13∣
b
tan-1(3/2)
c
tan-1(4/9)
d
sin-1(4/9)
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
H=u2sin2θ2 g or 80=u2sin2θ2×10 u2sin2θ=1600 or usinθ=40 ms-1 Horizontal velocity =ucosθ=3×30=90 ms-1 usinθucosθ=4090 tanθ=49 θ=tan-149