Circular MotionMedium
Question
A stone is projected from level ground at t = 0 sec such that its horizontal and vertical components of initial velocity are 10 m/s and 20 m/s respectively. Then the instant of time at which magnitude of tangential and magnitude of normal components of acceleration of stone are same is : (neglect air resistance) g = 10 m/s2.
Options
A.1/2 sec
B.1 sec
C.3 sec
D.4 sec
Solution
Tangential acceleration = at = g sin θ
Normal acceleration = an = g cos θ
at = an
g sin θ = g cos θ ⇒ θ = 45o
⇒ vy = vx
uy − gt = ux
20 − (10)t = 10
t = 1 sec.
During downward motion
at = an
vy = – vx
20 − 10 t = − 10 ⇒ t = 3 sec.
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