Question

PART C (6pts)Projectile motionChallenge: Set the target at 15m away. Find two groups of values of initial velocity and initialangle that will hit the target with three stars!Take a screenshot and paste it here,

Answer 1

Answer:

v = 12.12 m/s and θ = 45°

v = 17.15 m/s and θ = 15°

Explanation:

The range in projectile motion is calculated as:

$R=\frac{v^2\sin2\theta}{g}$

Where v is the initial velocity and θ is the initial angle.

In this case, R = 15 m, and we want to find values of v and θ that make R = 15 m. To do this, we need to assign values to one of the variables and solve for R.

If θ = 45°, then

$\begin{gathered} 15=\frac{v^2\sin(2\cdot45)}{9.8} \\ \\ 15\cdot9.8=v^2\sin(90) \\ 147=v^2\cdot1 \\ 147=v^2 \\ \sqrt{147}=v \\ 12.12=v \end{gathered}$

Therefore, the group v = 12.12 m/s and θ = 45° hits the target.

In the same way, If θ = 15°, we get

$\begin{gathered} 15=\frac{v^2\sin(2\cdot15)}{9.8} \\ \\ 15\cdot9.8=v^2\sin(30) \\ 147=v^2(0.5) \\ \\ \frac{147}{0.5}=v^2 \\ \\ 294=v^2 \\ \sqrt{294}=v \\ 17.15=v \end{gathered}$

Therefore, the groups v = 17.15 m/s and θ = 15° hits the target.

So, the answers are

v = 12.12 m/s and θ = 45°

v = 17.15 m/s and θ = 15°