Asked by tim
At what projection angle will the range of a projectile equal to 6 times its maximum height?
I know its easy but ive been absent and missed class!
I know its easy but ive been absent and missed class!
Answers
Answered by
bobpursley
range=VcosTheta*time
height=VsinTheta*time-4.9t^2
or
max height occurs when Vv=0
Vvf=0=VsinTheta-9.8t
solve for t.
then put it in
hmax=VsinTheta*t-4.9t^2
solve for max height
Then, multiply it by six, and set it equal to VcosTheta*2*time (twice the time to the max height).
Prepare yourself for messy algebra.
height=VsinTheta*time-4.9t^2
or
max height occurs when Vv=0
Vvf=0=VsinTheta-9.8t
solve for t.
then put it in
hmax=VsinTheta*t-4.9t^2
solve for max height
Then, multiply it by six, and set it equal to VcosTheta*2*time (twice the time to the max height).
Prepare yourself for messy algebra.
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