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Qwerty123!
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Homework Statement
a projectile is launched at an angle of 26.5 above the horizontal at a height of 1.11m. the range is 2.331m. find the initial velocity and time given that acceleration due to gravity is 9.8m/s^2.
The Attempt at a Solution
first, I split the velocity into its components:
v1x = v1cos26.5
v1y = v1cos26.5
then,
v1x = dx/t
v1cos26.5 = dx/t
rearranged to isolate for time since it is not given:
t = dx/ v1cos26.5
also, using the kinematic equation: v2y^2 = v1y^2 + 2at
and since at the projectile's max, v2y = 0m/s and since the max occurs at t*(0.5) ***(but I think this is inaccurate because the projectile does not land at the same height in which it was released)
so,
v2y^2 = v1y^2 + 2at
0 = v1y^2 + 2(9.8)(t/2)
then isolate for t again,
t = 2(v1y)^2 / 4.9
or
t = 2(v1sin26.5)^2 / 4.9
then make each isolated time equation equal one another:
dx/v1cos26.5 = 2(v1sin26.5)^2 / 4.9
then isolate for v1:
v1 = sqrt( 4.9*dx / (sin26.5cos26.5) )
since dx = 2.331,
v1 = sqrt( 4.9*2.331/ (sin26.5cos26.5) )
v1= 5.35m/s
Therefore, v1 = 5.35m/s
then for time,
v1x = dx/t
t = dx/v1cos26.5
= 2.331 / 5.35cos(26.5)
= 0.487s
Therefore, t = 0.49s