Monoxdifly
MHB
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Something was thrown with an elevation angle $$60^{\circ}$$. Determine its velocity and direction after:
a. $$\frac{1}{2}\sqrt3$$ second
b. $$\sqrt3$$ second
I tried solving the (a) question by substituting the angle to the equation $$y=v_0\sin\alpha t-\frac{1}{2}gt^2$$ and got 0 (the book uses $$10m/s^2$$ as the gravity acceleration), which is impossible since at speed 0 the thing must have been stopped, but when I determined the time when it reached the maximum distance using the formula $$t_{\text{x max}}=\frac{2v_0\sin\alpha}{g}$$ I got $$\sqrt3$$ second which means the thing hasn't stopped yet in $$\frac{1}{2}\sqrt3$$ second. Did I make a mistake somewhere?
a. $$\frac{1}{2}\sqrt3$$ second
b. $$\sqrt3$$ second
I tried solving the (a) question by substituting the angle to the equation $$y=v_0\sin\alpha t-\frac{1}{2}gt^2$$ and got 0 (the book uses $$10m/s^2$$ as the gravity acceleration), which is impossible since at speed 0 the thing must have been stopped, but when I determined the time when it reached the maximum distance using the formula $$t_{\text{x max}}=\frac{2v_0\sin\alpha}{g}$$ I got $$\sqrt3$$ second which means the thing hasn't stopped yet in $$\frac{1}{2}\sqrt3$$ second. Did I make a mistake somewhere?