Question

A projectile is launched from the ground, such that it hits a target on the ground which is \(90\,\text{m}\) away. The minimum velocity of projectile to hit the target is \((\text{acceleration due to gravity}=10\,\text{m s}^{-2})\)

• A. \(10\,\text{m s}^{-1}\)
• B. \(16\,\text{m s}^{-1}\)
• C. \(60\,\text{m s}^{-1}\)
• D. \(30\,\text{m s}^{-1}\)

Hint:

For maximum range with minimum speed on level ground, use \[ R_{\max}=\frac{u^2}{g} \] which occurs when \(\theta=45^\circ\).

Answer 1

The Correct Option is D

Step 1: Recall the range formula for projectile motion.
For a projectile launched at angle $\theta$ with speed $u$ from the ground and landing at the same level, the horizontal range is: \[ R = \frac{u^2 \sin 2\theta}{g} \] Here $R = 90$ m and $g = 10$ m/s^2.
Step 2: Identify the condition for minimum launch speed.
For a fixed range $R$, the required speed $u$ depends on the angle $\theta$. From the range formula: \[ u^2 = \frac{Rg}{\sin 2\theta} \] The speed $u$ is minimum when $\sin 2\theta$ is maximum, i.e., $\sin 2\theta = 1$, which occurs at $\theta = 45^\circ$.
Step 3: Substitute to find minimum speed.
With $\sin 2\theta = 1$: \[ u^2 = Rg = 90 \times 10 = 900 \] \[ u = \sqrt{900} = 30 \text{ m/s} \]
Step 4: Check the logic of the angle.
At $\theta = 45^\circ$, $\sin 90^\circ = 1$ is the maximum value. Any other angle requires more speed for the same range because the sine factor would be less than 1. So $45^\circ$ indeed gives minimum speed.
Step 5: Verify with units and numbers.
$u^2 = 90 \times 10 = 900$ m^2/s^2, $u = 30$ m/s. This matches option 4.
Step 6: State the conclusion.
The minimum velocity to hit the target 90 m away is 30 m/s, launched at $45^\circ$ to the horizontal. \[ \boxed{30 \text{ m/s}} \]