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Physics Question on Motion in a straight line

A ball is projected horizontally with a velocity of 4 ms−14\,m s^{-1} from the top of a tower. The velocity of the ball after 0.7 s is (Take g = 10 ms−2m s^{-2})

A

1 ms−11\,ms^{-1}

B

10 ms−110\, ms^{-1}

C

8 ms−18\,ms^{-1}

D

3 ms−13 \,ms^{-1}

Answer

8 ms−18\,ms^{-1}

Explanation

Solution

Velocity of the ball after time t
=vx2+vy2=\sqrt{v^2_x+v^2_y}
=(4)2+(gt)2=\sqrt{(4)^2+(gt)^2}
=16+(10×0.6)2=\sqrt{16+(10 \times 0.6)^2}
= 8ms−18 ms^{-1}