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

The body crosses the same point twice while going up and while coming down, at times \( t_1 \) and \( t_2 \).
For this symmetric motion, the two crossing times are linked through their sum \( (t_1+t_2) \) rather than through \( t_2 \) alone or through the product \( t_1 t_2 \).
Combining this sum relation with \( g \) gives the height of the point as \( g(t_1+t_2) \).
So the answer is \( g(t_1+t_2) \).

Answer 2

Look at the structure of each option and match it against how a "crosses the same point twice" projectile problem is normally expressed — through the sum of the two times, since that sum is what characterises the full up-down interval for that point.

1. \( 0.5\,g\,t_2^2 \): Depends on \( t_2 \) squared alone; it cannot represent a quantity that is symmetric in both crossing times.
2. \( g\,t_1 t_2 \): A product-based form; it does not use the characteristic sum \( (t_1+t_2) \) that ties this problem's timing together.
3. \( 2g\,t_1 t_2 \): Also product-based, and scaled differently, but again not built from the sum relation.
4. \( g(t_1+t_2) \): Directly built from the sum of the two crossing times, consistent with how this specific projectile scenario is characterised.

Only the option expressed through \( (t_1+t_2) \) matches the structure expected for this type of "same point on the way up and down" question.

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