Question

# Two bodies are thrown simultaneously from a tower with same initial velocity $${v_0}$$ one vertically upwards, the other vertically downwards. The distance between the two bodies after time t is.

A
2v0t+12gt2
B
2v0t
C
v0t+12gt2
D
v0t

Solution

## The correct option is B $$2{v_0}t$$For vertically upward motion,$$h_1=v_0t-\cfrac{1}{2}gt^2$$and for vertically downward motion,$$h_0=v_0t+\cfrac{1}{2}gt^2$$$$\therefore$$ Total distance covered in $$t$$ sec$$h=h_1+h_2=2v_0t$$Physics

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