- #1

Monoxdifly

MHB

- 284

- 0

a. \(\displaystyle \frac{1}{2}\sqrt3\) second

b. \(\displaystyle \sqrt3\) second

I tried solving the (a) question by substituting the angle to the equation \(\displaystyle y=v_0\sin\alpha t-\frac{1}{2}gt^2\) and got 0 (the book uses \(\displaystyle 10m/s^2\) 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 \(\displaystyle t_{\text{x max}}=\frac{2v_0\sin\alpha}{g}\) I got \(\displaystyle \sqrt3\) second which means the thing hasn't stopped yet in \(\displaystyle \frac{1}{2}\sqrt3\) second. Did I make a mistake somewhere?