At t = 0, an arrow is fired vertically upwards with a speed of 100 ms-1. A second arrow is fired vertically upwards with the same speed at t = 5 s. Then

Let they meet at height h after time t.h=100t−12gt2→ for first arrow =100(t−5)−12g(t−5)2→ for second arrow ⇒ t=12.5s (after solving). So (1) is correct. Time of flight of first arrow: T=2ug=2×10010=20sSecond arrow will reach after 5 s of reaching first. So (2) is correct. v1=100−10×20=−100ms−1v2=100−10×15=−50ms−1 Ratio: v1v2=2:1.So(3) is correct. Maximum height attainedH=u22g=(100)22×10=500m. Hence, (4) is incorrect.