   Chapter 12.3, Problem 35E

Chapter
Section
Textbook Problem

# A shot fired from a gun with a muzzle speed of 1200 feet per second is to hit a target 3000 feet away. Determine the minimum angle of elevation of the gun.

To determine

To calculate: The angle of elevation of the gun whose range is r(t)=3000ft, nozzle velocity v0=1200ft/s and g=32 ft/sec2.

Explanation

Given:

Range of the gun is r(t)=3000ft.

Nozzle velocity of the gun is v0=1200ft/s.

And g=32 ft/sec2.

Formula used:

The path followed by a projectile is described as:

r(t)=(v0cosθ)t i+[(v0sinθ)t+12gt2]j

Calculation:

The range of the shot fired is given by:

r(t)=(v0cosθ)t i+[h+(v0sinθ)t16t2]j

A shot is fired from the gun in a horizontal plane, so the height (h=0), and the range is given as 3000ft.

Thus, substitute the values as follows and obtain:

r(t)=(v0cosθ)t i+[(v0sinθ)t16t2]j3000=(v0cosθ)t i+[(v0sinθ)t16t2]j

This gives:

x=(v0cosθ)t=3000 …… (1)

y=(v0

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