Question

Two object, A and B vertical thrown up with velocities, 80ms^-2 and 100ms^-2 at 2 sec intervals. Where and when will the two objects meet each other. (G = 10ms^-2)

Answer 1

Answer:

2

Explanation:

Answer 2

The two objects will meet at t = 4.5 seconds and at height of 288.75m above the ground.

Inorder to find the time at which the two objects will meet,

Since at the time they meet their displacement from the ground are same.

Therefore,

• s1 = u1t + 0.5at1^2 = s2 = u2t + 0.5a t2^2

We know B is thrown 2 seconds after A.

• Therefore t2  = t1 - 2 = t -2

Also u1 = 80m/s

and u2 = 100m/s

a = -g = -10m/s^2

• Therefore we  can equate the displacement of A and B
• 80t - 0.5 x 10 xt^2 = 100 (t - 2) - 0.5x10x(t - 2)^2

• ==> 20t - 200 + 5t^2  - 5x(t^2 - 4t + 4) = 0

• ==> 20t - 200 + 5t^2 - 5t^2 + 20t -20 = 0

• ==> 40t = 220

• ==> t = 5.5sec

At 5.5 sec their displacement is,

• s = 80x4.5 - 5(4.5)^2 = 440 - 151.25 = 288.75m

Checking with the second ball,

• s = 100( 3.5) - 5 (3.5)^2 = 350 - 61.25 = 288.75m

Therefore, the two objects A and B will meet at t = 4.5 seconds and at height of 288.75m above the ground.