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?
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
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
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
To determine the time it takes for the two bodies to meet, we need to consider their respective motion and find the point where their paths intersect.
Let's assume that the first body is thrown at time t = 0, and the second body is thrown at time t = 4 seconds. Since both bodies are thrown vertically upwards with an initial velocity of 98 m/s, we can use the kinematic equation for vertical motion:
s = ut - (1/2)gt^2
where:
s = displacement
u = initial velocity
t = time
g = acceleration due to gravity (approximately 9.8 m/s^2)
For the first body:
s1 = (98t1) - (1/2)(9.8)(t1)^2
For the second body:
s2 = (98(t2 - 4)) - (1/2)(9.8)(t2 - 4)^2
Since the two bodies meet at some point, their displacements will be equal. Therefore, we can set up an equation:
(98t1) - (1/2)(9.8)(t1)^2 = (98(t2 - 4)) - (1/2)(9.8)(t2 - 4)^2
Simplifying the equation, we have:
98t1 - 4.9(t1)^2 = 98t2 - 392 - 4.9(t2^2 - 8t2 + 16)
Rearranging and collecting the like terms, we get:
4.9(t1^2 - t2^2) + 4.9(8t2 - t1) - 392 = 0
This is a quadratic equation, which can be solved to find the values of t1 and t2. Once you solve the equation, you can determine the time after the first body is thrown when they will meet.