A body is projected from height of 60 m with a velocity 10 ms^-1 at angle 30⁰ to horizontal. The time of flight of the body is [g =10 ms^-2]

Time of Flight Solution

Given data:

• Initial velocity (\(u\)) = 10 m/s
• Angle of projection (\(\theta\)) = 30°
• Height (\(h\)) = 60 m
• Acceleration due to gravity (\(g\)) = 10 m/s²

Step 1: Components of velocity

- Vertical component of velocity: u_y = u sin θ = 10 sin 30° = 10 × 0.5 = 5 m/s

- Horizontal component of velocity: u_x = u cos θ = 10 cos 30° = 10 × √3/2 ≈ 8.66 m/s

Step 2: Time of flight calculation

The time of flight is determined by the vertical motion of the body. Using the kinematic equation:

y = u_y t + (1/2)(-g)t²

Here, \(y = -60\) (since the body falls 60 m below the starting height).

Substitute the values:

-60 = 5t - (1/2)(10)t²

-60 = 5t - 5t²

Rearranging:

5t² - 5t - 60 = 0

Simplify:

t² - t - 12 = 0

Step 3: Solve the quadratic equation

The quadratic equation is:

t² - t - 12 = 0

Factorizing:

(t - 4)(t + 3) = 0

Thus, \(t = 4 \, \text{s}\) (since time cannot be negative).

Correct option: (d) 4 s