A stone is projected from level ground at t=0 sec such that its horizontal and vertical components of initial velocity are 10m/s and 20m/s respec ,then the instant of time at which magnitude of tangential and magnitude of normal components of acceleration of stone are sameis
Asked by m.nilu
22nd August 2018,
12:37 PM
Answered by Expert
Answer:
horizontal component of projected velocity = ucosθ = 10 m/s
vertical comonent of projected velocity = usinθ = 20 m/s
hence angle of projection, θ = tan-1(2) = 63º
After some time normal component of acceleration ( g×sinφ ) equals the
tangential component of acceleration ( g×cosφ ), where φ is the angle made by the velocity vector with horizontal.
g×sinφ = g×cosφ or φ = 45º
when the velocity vector makes an angle 45º to horizontal, then vertical and horizontal components are same,
hence vertical component of velocity is now 10 m/s.
time of this instant t is calcuted from, 10 = 20 - g×t ...........(1)
if we assume g = 10 m/s2 , then we get from eqn.(1) , t = 1 s
Answered by Expert
23rd August 2018,
1:12 PM
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