A vertically projected particle reaches maximum height H. The ratio of its first (ascending) and second (descending) times at height h<H is 1/3. Find 3H/h.
Physics · Mixed Chapter
JEE Main Practice

Question

A vertically projected particle reaches maximum height H. The ratio of its first (ascending) and second (descending) times at height h<H is 1/3. Find 3H/h.

G1 Physics · Accepted Answer

Step 1: Times at height h satisfy h=ut−½gt², giving t₁,₂=(u∓√(u²−2gh))/g. Step 2: Set t₁/t₂=(u−s)/(u+s)=1/3, where s=√(u²−2gh); solve 3u−3s=u+s⇒s=u/2. Step 3: u²−2gh=u²/4 ⇒h=3u²/(8g). Step 4: Since H=u²/(2g), h=3H/4; therefore 3H/h=4.