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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