Question

An object projected vertically up from the top of the tower took $5$ s to reach the ground. The average velocity of the object is $5{ms}^{-1})$, find its average speed. (given $g=10{ms}^{-1})$.

• A. $65{ms}^{-1}$
• B. $13{ms}^{-1}$
• C. $26{ms}^{-1}$
• D. $25{ms}^{-1}$

Answer 1

The correct option is B $13{ms}^{-1}$

Height of tower= displacement of an object= average velocity $\times$ time$=5\times 5=25m$,

Let initial velocity of the object is $u$.

Using $s=ut+\frac{1}{2}a{t}^2$, $\Rightarrow -25=u\times 5-\frac{1}{2}g\times 5^2$ $\Rightarrow u=20m{s}^{-1}$,

Now from the tower,Let maximum height reached by object is $h$.

At maximum height velocity $v=0$,

Using $v^2=u^2+2as$ $\Rightarrow 0={20}^2-2gh$ $\Rightarrow h=20m$,

$\therefore$ Total distance $=2h+25=65m$

$\text{Average speed}=\frac{\text{Total Distance}}{\text{Total Time}}$,

$\Rightarrow$ Average speed$=\frac{65}{5}=13m{s}^{-1}$