Question

The magnitude of the ball's velocity just before it hits the ground is $166\times {10}^{-x}m{s}^{-1}$. Find the value of x.

Answer 1

At height $h=9.1m$, the magnitude of velocity, $v^2={7.6}^2+{6.1}^2=94.97$ and this is the intial velocity when the ball fall down.

Using the formula:

$v^2-u^2=2aS\Rightarrow v^{\prime 2}-\left(94.97\right)=2\left(10\right)\left(9.1\right)$

$\Rightarrow v^{\prime }=16.6m/s$

$\Rightarrow x=1$