A piece of marble is projected from the earth's surface with a velocity of `50 m s^(-1)` . 2 s later , it just clears a wall 5 m high. What is the angle of projection ?

Options

A. `45^@`
B. `30^@`
C. `60^@`
D. `sin ^(-1)(1/4)`

Answer

To solve the problem step by step, we will use the equations of motion for projectile motion.

**Step 1: Identify the given values.**
- Initial velocity (u) = 50 m/s
- Height of the wall (h) = 5 m
- Time taken to reach the wall (t) = 2 s
- Acceleration due to gravity (g) = 10 m/s² (approximately)

**Step 2: Break down the initial velocity into components.**
- The horizontal component of the initial velocity: \( u_x = u \cos \theta \)
- The vertical component of the initial velocity: \( u_y = u \sin \theta \)

**Step 3: Use the vertical motion equation to find the angle.**
The vertical displacement (height) can be described by the equation of motion:
\[ h = u_y t - \frac{1}{2} g t^2 \]
Substituting the known values:
\[ 5 = (50 \sin \theta)(2) - \frac{1}{2} (10)(2^2) \]

**Step 4: Simplify the equation.**
Calculate the second term:
\[ \frac{1}{2} (10)(2^2) = \frac{1}{2} (10)(4) = 20 \]
So, the equation becomes:
\[ 5 = 100 \sin \theta - 20 \]
Rearranging gives:
\[ 100 \sin \theta = 25 \]

**Step 5: Solve for \( \sin \theta \).**
\[ \sin \theta = \frac{25}{100} = \frac{1}{4} \]

**Step 6: Find the angle \( \theta \).**
To find \( \theta \), we take the inverse sine:
\[ \theta = \sin^{-1}\left(\frac{1}{4}\right) \]

**Final Answer:**
The angle of projection \( \theta \) is \( \sin^{-1}\left(\frac{1}{4}\right) \).

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