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I need to know the speed and angle of flight of an arrow 3.35 seconds after it has been shot. The information: You are standing on a 30.0m tall cliff and shoot an arrow at 25.0 m/s at an initial angle of 32.0 degrees. What will be the speed and angle of the flight at 3.35 seconds?

Hello Seth,

I'll assume that the 32.0 degree angle is above horizontal. The vertical component of the initial 25.0 m/s is

Viv = Vi*sin32.0

To find, Vfv, the vertical component of velocity after 3.35 s, use the kinematic formula

Vfv = Viv + a*t

where a = -g = -9.8 m/s^2. Vfv is the magnitude of the vertical component of velocity.

The horizontal component of velocity is constant over time. (If we can assume no air resistance -- we have to assume that since the question didn't give us any details to calculate air resistance.) The horizontal component is

Vh = Vi*cos32.0

You have the vertical and horizontal component of velocity at t=3.35 s. The resultant of them is the hypotenuse of a triangle where the 2 components are 2 of the sides. Use Pythagoras to find the magnitude of the resultant. Find the angle using trigonometry -- like arctan(Vfv/Vh).

I hope this helps,

Steve

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I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.**Education/Credentials**

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