Question

A 9.40 kgprojectile is fired vertically upward. Air drag decreases the mechanical energy of the projectile-Earth system by 68.0 kJduring the projectile’s ascent. How much higher would the projectile have gone were air drag negligible?

Short answer

The height is h=738 m .

Step-by-step solution

Given data:

Mass of projectile, m = 9.40 kg

Decrease in mechanical energy, PE=68.0 kJ=68000 J

To understand the concept:

Potential energy is due to the position of the object from the earth’s surface.

Formula:

The potential energy is define by using following formula.

$PE=mgh$

Here, m is the mass, g is the acceleration due to gravity having a value $9.8m/s^2$, and h is the height.

Calculate the height:

Potential energy is given by,

$$\begin{gathered} PE=mgh \\ 68000J=9.40kg\times 9.8m/s^2\times h \\ h=738m \end{gathered}$$

Hence, 738 m much higher would the projectile have gone were air drag negligible.