Question

A body is projected vertically up at t = 0 with a



velocity of 98 m/s. Another body is projected from



the same point with same velocity after 4 seconds.



both bodies will meet at t =



(1) 6 (2) 8 s (3) 10 s (4) 12

Answer 1

Answer:

(4) 12

Explanation:

Let they meet after time t from projection of 1st at height h above the ground

$ut - 1 \div 2 \times g {t}^{2}$

For 1st h =98t-1/2×9.8×t^2━━━━━━━━━━eq.1

For 2nd h =98(t-4)-1/2×9.8×(t-4)^2━━━━━━eq.2

From eq. 1st and 2nd

0=98×4+9.8/2(-8t+16)

40-4t+8=0

4t=48

t=12s