A golf ball is hit from the ground with an initial velocity of 208 feet per second. Assume the starting height of the ball is 0 feet. How long will it take the golf ball to hit the ground
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It depends on the velocity (you've just given speed, which is the magnitude of velocity), but, given the initial conditions, the equation of motion in the vertical would be\[y=-16t^2+208 \sin \theta t\](which I can explain if you need). The maximal time of flight is had when the height of the ball is 0 (i.e. at the end). So we have from the quadratic:\[0=-16t^2+208 \sin \theta t \rightarrow t^2(t-13 \sin \theta ) =0\]which means either\[t=0\](i.e. beginning, which we're not interested in) or,\[t=13 \sin \theta\]The maximum value sine can take is 1, so the maximum time you can have is t=13, BUT you may also have t=6 for an angle of approx. 27.5 degrees. You need the initial direction of the ball.