Hard
Question
A point mass is projected, making an acute angle with the horizontal. If angle between velocity 
and acceleration 
is θ at any time t during the motion, then θ is given by
and acceleration is θ at any time t during the motion, then θ is given byOptions
A.0o < θ < 90o
B.θ = 90o
C.θ < 90o
D.0o < θ < 180o
Solution
Acute Angle of Velocity with horizontal possible is − 90o to + 90o hence angle with g is 0o to180o.
θ1 is acute
⇒ 0o ≤ θ1 < 90o (during the upward journey of mass)
from fig . θ = 90o + θ1
or, 90o ≤ θ < 180o .......(1)
During downward motion
0o < θ2 < 90o
θ = 90º − θ2
0o < θ < 90o .......(2)
From eq. (1) and (2)
i.e., 0 < θ < 90o U 90o ≤ θ < 180o
⇒ 0o < θ < 180o ″D″ Ans.
θ1 is acute
⇒ 0o ≤ θ1 < 90o (during the upward journey of mass)
from fig . θ = 90o + θ1
or, 90o ≤ θ < 180o .......(1)
During downward motion
0o < θ2 < 90o
θ = 90º − θ2
0o < θ < 90o .......(2)
From eq. (1) and (2)
i.e., 0 < θ < 90o U 90o ≤ θ < 180o
⇒ 0o < θ < 180o ″D″ Ans.
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