Question

21
Two bodies are thrown vertically upward with same initial velocity of 100 m/s, from same point but 4s apa
The distance moved by the two bodies by the time they meet would be and respectively. [g=18
m/s?]

Answer 1

Answer:

Both the bodies meet each other at t secs after the first one is thrown

Displacement covered by first object in t secs s1=ut−21gt2=98t−219.8×t2

Displacement covered by second object in (t-4) secs s2=98(t−4)−219.8×(t−4)2

Since they meet s1=s2

We get t=12secs

Explanation:

mark me as brainleist