The problem in the image involves projectile motion, where Yusuf needs to shoot a pellet at a target located 10 meters away, with an initial velocity of $v_0 = 150 \, \text{m/s}$. The goal is to determine the angle $\theta$ at which Yusuf should aim above the horizontal to account for gravity.

Solution Outline:

1. Projectile motion basics:

   • The horizontal range $R$ of a projectile is given by the formula: $R = \frac{v_0^2 \sin(2\theta)}{g}$ where:
     • $R$ is the horizontal distance to the target (10 m in this case),
     • $v_0$ is the initial velocity (150 m/s),
     • $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$),
     • $\theta$ is the angle above the horizontal.

2. Rearrange the formula to solve for $\theta$:

   • From the formula above, solve for $\theta$ by first isolating $\sin(2\theta)$: $\sin(2\theta) = \frac{R \cdot g}{v_0^2}$
   • Substitute the given values: $\sin(2\theta) = \frac{10 \cdot 9.81}{150^2}$ $\sin(2\theta) \approx \frac{98.1}{22500}$ $\sin(2\theta) \approx 0.00436$
   • Solve for $2\theta$: $2\theta = \sin^{-1}(0.00436)$ $2\theta \approx 0.2497^\circ$
   • Therefore: $\theta \approx \frac{0.2497^\circ}{2} = 0.1249^\circ$

Conclusion:

Yusuf should aim at an angle of approximately 0.125° above the horizontal in order to hit the target that is 10 meters away.

Would you like more details or have any questions about this calculation? Here are 5 additional questions that build on this concept:

1. How would the angle change if the distance to the target doubled?
2. What would the angle be if the initial velocity was reduced to 100 m/s?
3. Can you calculate the total time the pellet is in the air before it hits the target?
4. How would air resistance affect this calculation in a real-world scenario?
5. What would happen if the angle were higher than the calculated value?

Tip: When solving projectile motion problems, always consider breaking the motion into horizontal and vertical components for clearer calculations.