t = 39.7/37cos19.5
y = 37sin19.5 t - 1/2*9.8*t^2 +1.5
A third baseman makes a throw to first base 39.7 m away. The ball leaves his hand with a speed of 37.0 m/s at a height of 1.5 m from the ground and making an angle of 19.5 o with the horizontal. How high will the ball be when it gets to first base?
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