Question

if an object is dropped from an initial height ( h ), its velocity at impact with the ground is given by

( v=sqrt{2 g h} )

where ( g ) is the acceleration due to gravity and ( h ) is the initial height.

(a) find the initial height (in feet) of an object if its velocity at impact is ( 52 mathrm{ft} / mathrm{sec} ). (assume that the acceleration due to gravity nearest hundredth of a foot, if necessary.

(b) find the initial height (in meters) of an object if its velocity at impact is ( 34 mathrm{~m} / mathrm{sec} ). (assume that the acceleration due to gravi the nearest tenth of a meter.

part: ( 0 / 2 )

part 1 of 2

(a) the initial height of the object is approximately ( square mathrm{ft} ).

Explanation:

Step1: Recall the formula for velocity at impact

We know the formula is \( v = \sqrt{2gh} \). We need to solve for \( h \). First, square both sides of the equation to get rid of the square root: \( v^2 = 2gh \).

Step2: Solve for \( h \)

Rearrange the formula to solve for \( h \): \( h=\frac{v^2}{2g} \). For part (a), we use the acceleration due to gravity in feet per second squared, which is \( g = 32\space\text{ft/s}^2 \), and \( v = 52\space\text{ft/s} \).

Step3: Substitute the values into the formula

Substitute \( v = 52 \) and \( g = 32 \) into the formula for \( h \): \( h=\frac{52^2}{2\times32} \).

Step4: Calculate the numerator and denominator

First, calculate \( 52^2 = 2704 \). Then, calculate \( 2\times32 = 64 \). Now, divide \( 2704 \) by \( 64 \): \( h=\frac{2704}{64}=42.25 \).

Answer:

The initial height of the object is approximately \( \boldsymbol{42.25} \) ft.