Chapter 3, Problem 28P

To determine

To show: The speed of projectile leaves the ground is equal to speed of strike.

Explanation of Solution

Given:

The time of ascent is equal to time of descent and firing level and landing level are equal.

Formula used:

The equation of velocity

  $\begin{array}{l}{v}_f=v_i-at\\ v_f=\text{ final velocity}\\ v_i=\text{ initial velocity}\\ a=\text{ acceleration due to gravity}\\ t=\text{ time}\end{array}$

Calculation:

When the body is launched from the surface of earth it has maximum velocity. Let’s consider its velocity v_iand when body reaches maximum height its velocity become zero.

Substituting value in equation.

  $\begin{array}{l}0=v_i-at\\ v_i=at\mathrm{.................}\left(1\right)\end{array}$

When body starts moving from highest point to surface of earth, it velocity at highest point is zero and final velocity is maximum.

Substituting the value, in equation of velocity.

  $\begin{array}{l}{v}_f=0+at\\ v_f=at\mathrm{............}\left(2\right)\end{array}$

Comparing (1) and (2).

The velocity of firing equal to velocity of landing.

Conclusion:

Thus, the velocity of firing equal to velocity of landing.