Question

A bullet is fired from origin with speed 10 m/s. the bullet is required to hit a target at point (5, y). Then possible value(s) of y. Horizontal surface is on x-axis and vertically above has been taken as y-axis.

Answer 1

The bullet can hit a target at $(5, y)$ only if

$$0 \le y \le \frac{15}{4} \, \text{m}.$$

Answer 2

1. Write the projectile equation: $y=x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$.

2. Substitute $x=5$, $u=10$, $g=10$ to get $y = 5\tan\theta - \frac{5}{4}\sec^2\theta$.

3. Express in terms of $u=\tan\theta$: $y = -\frac{5}{4}u^2 + 5u -\frac{5}{4}$.

4. Find the vertex (maximum) at $u=2$ yielding $y_{\max}=\frac{15}{4}$.

5. With physical constraints (the projectile starting from ground and landing on ground), $y$ at $x=5$ varies from $0$ to $\frac{15}{4}$.