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/s 2 .
sec
Text Solution
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(b, c): Tangential acceleration = a t = g sin θ
Normal acceleration = a n = g cos θ
a t = a n
g sin θ = g cos θ ⇒ θ = 45°
⇒ v y = v x
u y – gt = u x
20 – (10)t = 10
t = 1 sec.
During downward motion
a t = a n
v y = – v x
20 – 10 t = – 10 ⇒ t = 3 sec.
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