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 ft/sec. (assume that the acceleration due to gravity is ( g=32 mathrm{ft} / mathrm{sec}^{2} ).) round to the nearest hundredth of a foot, if necessary. (b) find the initial height (in meters) of an object if its velocity at impact is 34 m/sec. (assume that the acceleration due to gravity is ( g=9.8 mathrm{~m} / mathrm{sec}^{2} ).) round to the nearest tenth of a meter. (a) the initial height of the object is approximately ( 42.25 mathrm{ft} ). (b) the initial height of the object is approximately ( square mathrm{m} ).

Explanation:

Part (b)

Step 1: Start with the velocity formula

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

Step 2: Solve for \( h \)

Divide both sides of the equation \( v^2 = 2gh \) by \( 2g \) to isolate \( h \). So, \( h=\frac{v^2}{2g} \).

Step 3: Substitute the given values

We are given \( v = 34 \, \text{m/sec} \) and \( g = 9.8 \, \text{m/sec}^2 \). Substitute these values into the formula for \( h \): \( h=\frac{34^2}{2\times9.8} \).

Step 4: Calculate the numerator and denominator

First, calculate \( 34^2 = 1156 \). Then, calculate \( 2\times9.8 = 19.6 \). Now, divide \( 1156 \) by \( 19.6 \): \( h=\frac{1156}{19.6}\approx58.97959\).

Step 5: Round to the nearest tenth

Rounding \( 58.97959 \) to the nearest tenth gives \( 59.0 \).

Answer:

\( 59.0 \)