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.
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...