Question

answer this it's urgent

At t=0, an arow is fired vertically upwards with a speed of 100ms1 A second arrow is fired vertically upwards with the same speed
at t= 5s. Then (g-1Om/s)
(A) the two arrows will be at the same height above the ground at t = 12.5s
(B) the two arrows will reach back their starting points at t = 20s and t= 25s
(C) the ratio of the speeds of the first and second arrows at t = 20s will be 2: 1
(D) the maximum height attained by either arrow will be 1000m
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Answer 1

Answer:

A::B::C

 Explanation:

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 (a) is correct.

Time of flight of first arrow: T=2ug=2×10010=20s

Second arrow will reach after 5s of reaching first. So b is correct.

(v1=100−10×20=−100ms−1),(v2=100−10×15=−50ms−1):}

Ratio: v1v2=2:1.So (c) is correct.

Maximum height attained

H=u22g=(100)22×10=500m. Hence, d incorrect.