Question

A body is projected vertically upwards at time \(t = 0\) and it is seen at a height \(H\) at time \(t_1\) and \(t_2\) second during its flight. The maximum height attained is (acceleration due to gravity = \(g\))

• A. \(\dfrac{g(t_2 - t_1)^2}{8}\)
• B. \(\dfrac{g(t_1 + t_2)^2}{4}\)
• C. \(\dfrac{g(t_1 + t_2)^2}{8}\)
• D. \(\dfrac{g(t_2 - t_1)^2}{4}\)

Hint:

If a body reaches the same height twice, the times \(t_1\) and \(t_2\) are roots of the same quadratic, and \(t_1+t_2 = \frac{2u}{g}\).

Answer 1

The Correct Option is B

Step 1: Write the vertical motion equation.
For vertical projection, displacement at time \(t\) is:
\[ H = ut - \frac{1}{2}gt^2 \]
Since the body is at height \(H\) at times \(t_1\) and \(t_2\),
\[ H = ut_1 - \frac{1}{2}gt_1^2 \] \[ H = ut_2 - \frac{1}{2}gt_2^2 \]
Step 2: Use property of quadratic equation.
The equation \(ut - \frac{1}{2}gt^2 - H = 0\) has roots \(t_1\) and \(t_2\).
So sum of roots is:
\[ t_1 + t_2 = \frac{u}{\frac{1}{2}g} = \frac{2u}{g} \]
Hence,
\[ u = \frac{g(t_1+t_2)}{2} \]
Step 3: Maximum height attained.
Maximum height is:
\[ H_{\max} = \frac{u^2}{2g} \]
Substitute \(u\):
\[ H_{\max} = \frac{\left(\frac{g(t_1+t_2)}{2}\right)^2}{2g} \]
\[ H_{\max} = \frac{g(t_1+t_2)^2}{8} \]
Step 4: Matching with options.
The derived expression corresponds to option (C), but given key says (B).
However, the correct physical derivation gives:
\[ \boxed{\frac{g(t_1+t_2)^2}{8}} \]
So, the answer key appears mismatched for Q2.
Final Answer: \[ \boxed{\dfrac{g(t_1+t_2)^2}{8}} \]