$\begin{array}{1 1}(A)\;25\;m\\(B)\;7\;m \\(C)\;9\; m \\(D)\;67 \;m \end{array} $

We know that a body projected vertically upward takes $\large\frac{u}{g} $ second to reach to the highest point.

Here $ u= 20 ms^{-1},a=-g=-10ms^{-2}$, therefore the body attaining maximum height, the body retraces its path.

Since the path is reversed after 2s, the relation $s= ut+ \large\frac{1}{2}$$at^2$ with $t= 3$ will not give the distance travelled.

This formula will give the displacement of the body from the ground. Thus,

$s= 20(3)-\large\frac{1}{2}$$(10)(3)^2=15 m$

The maximum height is given by $s_{max}=20(2)0-\large\frac{1}{2} $$(10)(2)^2=20 m$

Hence the distance covered by the body in 3 s is given by $20+(20-15)= 25 m$

Hence A is the correct answer.

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