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how do you compare the velocity of the projectile to the height?
how do you compare the velocity of the projectile to the height?
Answer:
first you need to break the velocity of the projectile to its vertial and horizontal components,
for the sake of arguement we assume that the projectile leaves with a velocity of V, and at and angle of theta 45.
the vertical components of the velocity would be Vy = V.sin(theta)
the horizontal component of the velocity would be Vx = V.cos(theta)
now height of the projectle is only dependent on the vertical component.
Vy = V.sin(theta)
and as per the equations of motion, if there is no wind/drag on the projectile, then the horizontal velocity will stay constant through out the projectiles whole journey, however the vertical component would be affected by gravity. hence it would slow down.
we can compute the vertical height be using the equation
v^2 = u^2 + 2as
where
u is the initial velocity which Vy = V.sin(theta)
v is the final velocity = 0 (as the projectile would stop
a = g = acceleration due to gravity to be -9.81N
s = distance
so
0 = V.sin(theta) + 2 (-9.81)(s)
-V.sin(theta) = -2(9.81)(s)
V.sin(theta) = 2(9.81)(s)
V.sin(theta)/2(9.81) = s
that is your vertical height.
now if you want to know the height of the projectile at twice the V.
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