1.

A stone is dropped into water from a bridge 44.1m above the water. Another stone is thrown vertically downward 1 s later. Both strike the water simultaneously. What was the initial speed of the second stone?

Answer»

SOLUTION :`t=sqrt((2xx44.1)/(9.8))s=sqrt(9)s=3s`,
`44.1=vxx2+(1)/(2)xx9.8xx2xx2`
or `2v=44.1-4.9xx4=24.5`
or `V=(24.5)/(2)MS^(-1)=12.25ms^(-1)`


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