Question

a project 50 gram is fired in air such that the horizontal range and vertical height are equal calculate the time of flight ​

Answer 1

The calculated values are approximately:

- Launch Angle (θ): 63.43 degrees

- Initial Velocity (u): 0.74 m/s

- Time of Flight (T): 0.15 s

Given:

- Mass of the projectile (m) = 50 grams = 0.05 kg

- Acceleration due to gravity (g) = 9.8 m/s²

We want to find the initial velocity (u), launch angle (θ), and time of flight (T) for a 50-gram projectile to achieve equal horizontal range and vertical height.

1. **Calculate Launch Angle (θ):**

`\[ \tan(\theta) = 2 \]`

`\[ \theta = \tan^(-1)(2) \]`

Using a calculator: θ ≈ 63.43 degrees

2. **Calculate Initial Velocity (u):**

`\[ u = \sqrt{(Rg)/(\sin(2\theta))} \]`

R = H (since we want equal horizontal range and vertical height)

`\[ u = \sqrt{(Hg)/(\sin(\theta))} \]`

Substitute values:

`\[ u = \sqrt{((0.05 kg * 9.8 m/s^2))/(\sin(63.43 degrees))} \]`

`\[ u \approx \sqrt{(0.49 N)/(0.891)} \]`

`\[ u \approx 0.74 m/s \]`

3. **Calculate Time of Flight (T):**

`\[ T = (2u\sin(\theta))/(g) \]`

Substitute values:

`\[ T = (2 * 0.74 m/s * \sin(63.43 degrees))/(9.8 m/s^2) \]`

`\[ T \approx (1.48 m/s)/(9.8 m/s^2) \]`

`\[ T \approx 0.15 s \]`

The question probable may be:

What are the initial velocity and launch angle needed for a 50-gram projectile fired in the air to achieve equal horizontal range and vertical height? Calculate the time of flight under these conditions.