Asked by Chris
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?
Answers
Answered by
drwls
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
"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
Answered by
RESONITE
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
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
Answered by
RESONITE
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
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