Question

a body a is projected upwards with velocity 98 metre per second square the body b is projected up with the same velocity but after 4 second both will meet after ??

please tell this by any shortcut method ​​

Answer 1

this is the methord but answer is not sure you better doubdouble check it

let time taken by body 1 as (t)..then

second body time is equal to (t-4) cause it started 4s later

• I make 2 equations using s=ut+1/2at^2
• then I equal 1st equation to 2nd equation cause they reach same heights

Hope you can understand

Answer 2

Answer:

12s

Explanation:

Velocity of body A = 98 m/s

Body B projected upwards with the same initial velocity

Since they have same initial velocity , so their heights will be equal , H_A = H_B . As we know that ,

H = ut - 1/2 gt²

Let the time taken by the body A to reach point A be t . And here , the body B thrown after 4s , so the time taken will be ( t - 4 ) .

 

⇒ ut - 1/2 gt² = u(t - 4) - 1/2 g (t - 4)²

⇒ ut - 1/2 gt² = ut - 4u - 1/2 ×g [t²-2(t)(4)+4²]

⇒ ut - 1/2gt² = ut - 4u - 1/2 g[t²-8t + 16]

⇒ ut - 1/2gt² = ut - 4u - 1/2 gt² - 1/2(-8gt) - 1/2(16g)

⇒ ut - 1/2gt² - ut + 1/2gt² = - 4u + 4gt + 8g

⇒ 0 = -4u + 4gt - 8g

⇒ 4u - 4gt + 8g = 0

⇒ 4(98) - 4(9.8)t + 8(9.8) = 0

⇒ 392 - 39.2t + 78.4 = 0

⇒ 470.4 = 39.2 t

⇒ t = 12s

Bodies will meet after 12s .