Question

A body is thrown vertically upward. It passes from a point at times \( t_1 \) and \( t_2 \), then the height of that point will be

• A. 0.5 g \( t_2^2 \)
• B. g \( t_1 t_2 \)
• C. 2 g \( t_1 t_2 \)
• D. g \( (t_1 + t_2) \)

Hint:

In vertical motion, the total displacement at a given time can be derived using the equation \( h = v_0 t - \frac{1}{2} g t^2 \).

Answer 1

The Correct Option is D

Step 1: Understanding the motion.
When a body is thrown vertically upward, the displacement equation for the height \( h \) is given by: \[ h = v_0 t - \frac{1}{2} g t^2 \] where \( v_0 \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( t \) is the time. We need to find the height difference when the body passes through the same point at times \( t_1 \) and \( t_2 \).
Step 2: Using the equation of motion.
At time \( t_1 \), the height is: \[ h_1 = v_0 t_1 - \frac{1}{2} g t_1^2 \] At time \( t_2 \), the height is: \[ h_2 = v_0 t_2 - \frac{1}{2} g t_2^2 \] The total height covered from the initial position to the point at times \( t_1 \) and \( t_2 \) is the sum of the two heights: \[ h = g (t_1 + t_2) \] Thus, the correct formula for the height at that point is \( g(t_1 + t_2) \).
Step 3: Conclusion.
The correct answer is (4) g \( (t_1 + t_2) \), which gives the height when the body passes through the same point at two different times.