A particle takes 9 s and 4 s respectively to reach foot of the building when projected vertically upwards and downwards respectively with same speed from top of the building. How much time (in s) particle will take to reach foot of the building when it is dropped from top of the building?
Physics · Mixed Chapter
JEE Mains

Question

A particle takes 9 s and 4 s respectively to reach foot of the building when projected vertically upwards and downwards respectively with same speed from top of the building. How much time (in s) particle will take to reach foot of the building when it is dropped from top of the building?

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Let H be building height and u be the common launch-speed magnitude. Take downward positive. For upward projection for 9 s: H=−9u+(1/2)g(9²). For downward projection for 4 s: H=4u+(1/2)g(4²). Step 2: Equate heights: −9u+40.5g=4u+8g ⇒13u=32.5g ⇒u=2.5g. Step 3: Height is H=4(2.5g)+8g=18g. For drop from rest, H=(1/2)gt². Step 4: (1/2)gt²=18g ⇒t²=36 ⇒t=6 s.