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**1. A third baseman makes a throw to first base 38.6 m away. The ball leaves his hand with a speed of 38.0 m/s at a height of 1.4 m from the ground and making an angle of 24.2 degrees with the horizontal. How high will the ball be when it gets to first base?**

**2. d=V0(t) + 1/2(g)(t^2)**

**3. 38.6= 38(t) + 1/2(-9.81)(t^2) using the quadratic equation I get 1.2 s for t. I'm not sure what to do from here.**