Edits to “Solution”, “Answer”

Valter edited
revision #22173 parent #12478 ← older
@@ -31,12 +31,12 @@Solution
\fbox{ $t=\sqrt{\frac{2D}{g}}$ } \tag{1}
$$
−b) Note that the expression $(1)$ does not include the value of the angle, so all balls will be dropped simultaneously. They will lie on a circle of radius $r = g t^2/2$, as shown in the animation
+b) Note that the expression $(1)$ does not include the value of the angle, so all balls will be dropped simultaneously. They will lie on a circle of diameter $r = g t^2/2$, as shown in the animation
#### Answer
a. $t = \sqrt{2D/g}$
−b. On a circle of radius $\frac{gt^{2}}{2}$ with top point $A$.
+b. On a circle of diameter $\frac{gt^{2}}{2}$ with top point $A$.
unchanged lines 3