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.

Answer 2

A body thrown vertically upward passes the same point twice — once going up, once coming back down — at times \( t_1 \) and \( t_2 \). We can reason about this using the symmetry of the up-and-down motion and the initial speed relation for this classic scenario, then check which option matches.

1. \( 0.5\,g\,t_2^2 \): This form only involves the later time \( t_2 \) and leaves out \( t_1 \) entirely; since the point is passed at both times, an expression describing it should naturally involve both \( t_1 \) and \( t_2 \) together, so this does not fit.
2. \( g\,t_1 t_2 \): This combines both times through a product, but it does not correspond to the standard relation used for the initial speed of this kind of symmetric up-down motion, which is built from the sum \( (t_1+t_2) \) rather than the product alone.
3. \( 2g\,t_1 t_2 \): Similar to option (2), this uses the product of the times rather than their sum, and does not match the known relation for this problem.
4. \( g(t_1+t_2) \): For a body thrown upward and passing the same point on the way up and on the way down, the launch speed and the two crossing times are tied together through their sum, \( (t_1+t_2) \), and combining this relation with \( g \) gives the height of that point in terms of \( g \) and \( (t_1+t_2) \), matching this option.

Since the physical set-up ties the two crossing times together through their sum rather than through \( t_2 \) alone or through their product, the expression built from \( (t_1+t_2) \) is the one that correctly reflects the geometry of this motion.

Therefore, the correct answer is \( g(t_1+t_2) \).