Question

A ball is thrown vertically upward from the ground it crosses a point at the height of 25 metre twice at an interval of 4 seconds the ball was thrown with the velocity of

Answer 1

$$\begin{gathered} Lettheballclimbstoaheightof\mathit{h}afterthe25mmark. \\ Totalheightattainedbytheball=H=25+h \\ Now, \\ theballwilltravelhupandfallhdownin4s \\ Thus, \\ fromthetoppointitwillfallhinhalfthattime,i.e.,in2s \\ andattoppoint{v}_{goingupjourney}=u_{forreturnjourney}=0 \\ Therefore, \\ h=u_{forreturnjourney}\times t+\frac{1}{2}g{t}^2 \\ takingg=10m{s}^{-2} \\ h=0+\frac{1}{2}\times 10\times 2^2 \\ h=20m \\ therefore, \\ H=25+20 \\ H=45m \\ now, \\ fromthirdequationofmotionforgoingup: \\ {v}^2=u^2-2gH \\ 0=u^2-2\times 10\times 45 \\ {u}^2=900 \\ u=30m{s}^{-1} \end{gathered}$$