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Q. A ball is dropped from the top of a tower of height $100\,m$ and at the same time another ball is projected vertically upwards from ground with a velocity $25 \, ms^{-1}$. Then the distance from the top of the tower, at which the two balls meet is

KEAMKEAM 2014Motion in a Straight Line

Solution:

$ t=\frac{\text { height }}{\text { speed }}= \frac{100}{25}=4\, s$
Fall, $h =u t+\frac{1}{2}\, g t^{2} $
$h =0+\frac{1}{2} \times 9.8 \times 4^{2} $
$h =8 \times 9.8=78.4 \,m $