The time taken by a projectile thrown with a velocity `v` `cm//"sec"` at an angle `alpha` with the horizontal to attain the maximum height is given by

Options

A. `v//g`
B. `v//sinalpha`
C. `vsinalpha//g`
D. `(vsinalpha)g`

Answer

To find the time taken by a projectile thrown with a velocity \( v \) at an angle \( \alpha \) with the horizontal to reach its maximum height, we can follow these steps:

### Step 1: Understand the Components of Velocity
When a projectile is thrown at an angle \( \alpha \), its initial velocity \( v \) can be broken down into two components:
- Horizontal component: \( v_x = v \cos \alpha \)
- Vertical component: \( v_y = v \sin \alpha \)

### Step 2: Analyze the Vertical Motion
At the maximum height, the vertical component of the velocity becomes zero. Thus, we can set up the equation for vertical motion using the following kinematic equation:
\[ v_{y} = v_{y0} + a_y t \]
Where:
- \( v_{y} \) is the final vertical velocity (0 at maximum height)
- \( v_{y0} \) is the initial vertical velocity (\( v \sin \alpha \))
- \( a_y \) is the acceleration due to gravity (which is negative, \( -g \))
- \( t \) is the time taken to reach maximum height

### Step 3: Substitute the Known Values
Substituting the known values into the equation:
\[ 0 = v \sin \alpha - g t \]

### Step 4: Solve for Time \( t \)
Rearranging the equation to solve for \( t \):
\[ g t = v \sin \alpha \]
\[ t = \frac{v \sin \alpha}{g} \]

### Conclusion
Thus, the time taken by the projectile to reach its maximum height is given by:
\[ t = \frac{v \sin \alpha}{g} \]