Question

Two bodies are thrown vertically upwards with the same initially velocity of 98metre/sec but 4 sec apart.How long after is thrown will they meet?

Answer 1

In the United States,
"physical education" is the class devoted to exercise, fitness and sports. Your question is about physics.

Starting at t=0 when the first body is thrown, the height is
Y1 = 98t - 4.9 t^2. The second body's height, measured from the same t, is
Y2 = 98(t-4) - 4.9 (t-4)^2 (t>4)

Set the two equal and solve for t.

98t - 4.9 t^2 = 98(t-4) - 4.9 (t-4)^2

0 = -392 +4.9(8t)-16*4.9
39.2 t = 392 + 78.4 = 470.4
t = 12.0 seconds

The height at that time, for both objects, is 470.4 m

Answer 2

Both the bodies meet each other at t secs after the first one is thrown
Displacement covered by the first object in t sec
1
​
=ut−
2
1
​
gt
2
=98t−
2
1
​
9.8×t
2

Displacement covered by second object in (t-4) secs s
2
​
=98(t−4)−
2
1
​
9.8×(t−4)
2

Since they meet s
1
​
=s
2
​

We get t=12secs

Answer 3

SORRY FOR SOME ISSUES ABOVE
kindly check this one ..........

#Both the bodies meet each other at t secs after the first one is thrown
Displacement covered by the first object in t secs s1=ut−1/2gt^2=98t−1/2*9.8×t^2
Displacement covered by the second object in (t-4) secs s2=98(t−4)−1/2*9.8×(t−4)^2
Since they meet s1=s2
We get t=12secs