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If a ball is thrown vertically upwards with speed u, the distance covered during the last t second of its ascent is

$\begin{array}{1 1}(A)\;(u+gt) t \\(B)\;ut\\(C)\;\frac{1}{2} gt^2 \\(D)\;ut-\frac {1}{2}gt^2\end{array}$

Can you answer this question?

$S_{max}=ut_0 -\large\frac{1}{2}$$gt_0^2 and S= u(t_0 -t) -\large\frac{1}{2}$$g(t_0 -t)^2$
$d= S_{max}-S=ut -\large\frac{1}{2}$$g [t_0^2-(t_0-t)^2] \quad= \large\frac{1}{2}$$gt^2$
Hence C is the correct answer.
answered Jul 18, 2014 by