Asked by nirav Nirav
Is it possible to find the displacement on the ground that a
projectile will undergo after reaching back to ground with following
parameters?
angle = 89 degrees
initial velocity = say 50 m/s
the projectile initiates from ground itself. that is initial point is
0m from the reference.
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UP 0 DOWN 0 0 3
3 Answers
62This one seems simple to me...
The same standard question of range!!! Is it different from what felt it is!?
V=50
Vy=50sin89
Time to reach the top
=Vy/g
Total Time of flight: t = 2.Vy/g
Horizontal component of velocity = Vx= 50cos89
Total Horizontal displacement = Vx.t
Just substitute values of sin 89 cos 89 u will be thru!
1Projectile motion is a very easy chapter.
I will give you teach you very easy way of solving it.
If the angle is 89 degrees.
U can write...
Speed=ucos89(i)+(usin89-gt)(j)
If we integrate this than we get...
Distance=ucos89t(i)=(usin89t-1/2 gt2)(j)
Now if the particle comes to the ground then the j component is 0.
So usin89t=1/2gt2
Get t and then substitute that t in ucos89t
So we get the horizontal distance travelled...