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## Q1(a) An object is shot at at 25 degree angle with a velocity of 5 m/s. What is the velocity in the X-direction? Y-direction? Vor= 5cos (25)=4.53s, Voj=5Sin25=2.11″3 (b) Now that you have calculated the initial velocity in the Y-direction we can find the time it takes the projectile to reach its maximum height. We cleverly recognize that at the top the velocity in the y direction is 0 so we can solve: v, = v, sin 0, – gt

I need to solve v(sub y)=v(sub 0)sin θ(sub 0)-gt to find the time that it takes the projectile to reach maximum height (so I believe I’m solving for t?). I have previously found v(sub 0y) in part (a), but I’m unsure what v(sub 0) and v(sub y) are. How do I find those variables?

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