Grade 11Mechanics
A body of mass m is thrown upwards with angle theta with velocity v .find its velocity after t sec
1 Answer
Vikas TU
Dear Student,
The Projectile takes after an illustrative direction. The vertical segment of its speed midpoints to zero in view of the symmetry of the parabola (we are intended to disregard the impact of air resistance). The level segment is consistent until the shot strikes the ground (once more, overlooking resistance).
No edge is given, however the edge can't be overlooked.
In Cartesian directions, if v=(vx,vy,vz)v=(vx,vy,vz) and vzvz is
the vertical part of vv, at that point the normal speed is
v2x+v2y−−−−−−√vx2+vy2. In polar directions, if ϕϕ is
the point of vv above even and r=|v|r=|v| is the extent
of vv (the underlying velocity), at that point the normal
speed is rcosϕrcosϕ.
In Cartesian directions, if v=(vx,vy,vz)v=(vx,vy,vz) and vzvz is
the vertical part of vv, at that point the normal speed is
v2x+v2y−−−−−−√vx2+vy2. In polar directions, if ϕϕ is
the point of vv above even and r=|v|r=|v| is the extent
of vv (the underlying velocity), at that point the normal
speed is rcosϕrcosϕ.
.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
Thapar University)
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