- #1
Mac5214
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1. the question is this:
a projectile's speed at launch is four times its speed at the top of the trajectory. What is the launch angle?
I am given no numerical values, but from my understanding gravity is a constant 9.8, and at the maximum height velocity should be equal to 0. 2.the equation to find the maximum height is Hmax=Vi^2sin^2(theta)/2G and the equation for3. since the whole problem is diriving a equation i am lost as to where to start. other than knowing that the angle for maximum range is 45 degrees and that i can use the inverse of tan to find a angle of projection, but i am unsure how to find an angle that gives four times the initial speed. sorry if the question is sort of vague, it is all i was given. and thank you for any help you may provide.
a projectile's speed at launch is four times its speed at the top of the trajectory. What is the launch angle?
I am given no numerical values, but from my understanding gravity is a constant 9.8, and at the maximum height velocity should be equal to 0. 2.the equation to find the maximum height is Hmax=Vi^2sin^2(theta)/2G and the equation for3. since the whole problem is diriving a equation i am lost as to where to start. other than knowing that the angle for maximum range is 45 degrees and that i can use the inverse of tan to find a angle of projection, but i am unsure how to find an angle that gives four times the initial speed. sorry if the question is sort of vague, it is all i was given. and thank you for any help you may provide.
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