Question

A body is thrown vertically upwards from A, the top of a tower, and reaches the ground in time t₁. If it is thrown vertically downwards from A with the same speed, it reaches the ground in time t₂. If it is allowed to fall freely from A, then the time taken to reach the ground is

• A. \( t = \dfrac{t_1 + t_2}{2} \)
• B. \( t = \dfrac{t_1 - t_2}{2} \)
• C. \( t = \sqrt{t_1 t_2} \)
• D. t = √((t₁)/(t₂))

Hint:

For problems involving throws from the same height: • Use equations of motion consistently. • Free-fall time often comes as the geometric mean.

Answer 1

The Correct Option is C

Step 1: Let height of tower be h and initial speed be u. Upward throw: h = ut₁ - (1)/(2)gt₁² ⋯ (1) Downward throw: h = ut₂ + (1)/(2)gt₂² ⋯ (2)
Step 2: Adding (1) and (2): 2h = u(t₁ + t₂) + (1)/(2)g(t₂² - t₁²) Subtracting (1) and (2): 0 = u(t₁ - t₂) - (1)/(2)g(t₁² + t₂²)
Step 3: Eliminating u and solving gives t = √(t₁ t₂)