Question

suppose that a falling body hits the group with velocity - 100ft/sec what was the initial height of the body?

Explanation:

Step1: Recall the velocity - height formula

The velocity - height formula for a freely - falling body is $v^{2}=v_{0}^{2}+2a(y - y_{0})$. In the case of a body falling near the surface of the Earth, the acceleration $a=- 32\mathrm{ft/s^{2}}$ (negative because it acts in the downward direction), the initial velocity $v_{0} = 0\mathrm{ft/s}$ (assuming it is dropped), the final velocity $v=-100\mathrm{ft/s}$ (negative because the direction is downward), and we want to find the initial height $y_{0}$ with respect to the ground level $y = 0$.

Step2: Substitute the values into the formula

Substituting $v_{0}=0$, $a=-32\mathrm{ft/s^{2}}$, $v = - 100\mathrm{ft/s}$ and $y = 0$ into $v^{2}=v_{0}^{2}+2a(y - y_{0})$, we get $(-100)^{2}=0^{2}+2\times(-32)\times(0 - y_{0})$.

Step3: Simplify the equation

$10000=64y_{0}$.

Step4: Solve for $y_{0}$

$y_{0}=\frac{10000}{64}=\frac{625}{4}=156.25\mathrm{ft}$.

Answer:

$156.25\mathrm{ft}$