First slide

Vertical projection from ground

Question

Two bodies are thrown upwards with same initial velocity $40 {ms}^{- 1}$ with time gap 4 sec. After what time from projection of first body they will meet $(g = 10 m / s^{2})$

Moderate

A

5 sec
B

6 sec
C

8 sec
D

2 sec

Solution

$Displacement covered by first object in t secs, s_{1} = ut - \frac{1}{2} {gt}^{2} = 40 t - \frac{1}{2} 10 \times t^{2} Displacement covered by second object in (t - 4) secs, s_{2} = 40 (t - 4) - \frac{1}{2} 10 \times (t - 4)^{2} Since they meet s_{1} = s_{2} We get t = 6 secs$

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