   Chapter 12, Problem 3PS

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# Projectile Motion A bomber is flying horizontally at an altitude of 3200 feet with a speed of 400 feet per second when it releases a bomb. A projectile is launched 5 seconds later from a cannon at a site facing the bomber and 5000 feet from the point that was directly beneath the bomber when the bomb was released, as shown in the figure. The projectile is to intercept the bomb at an altitude of 1600 feet. Determine the required initial speed and angle of inclination of the projectile. (Ignore air resistance.) To determine

To calculate: The required initial speed and angle of inclination of projectilethat intercepts the bomb at an altitude of 1600 feet. The bomber flying horizontally releases the bomb at 3200 ft altitude with a speed of 400 feet per second. The projectile is launched after 5 second as shown in the figure Explanation

Given: A bomber is flying horizontally at an altitude of 3200 ft with a speed of 400 feet per second. A projectile is launched after 5 second. The projectile intercepts the bomb at an altitude of 1600 feet. The given figure is,

Formula used: The vector position of projectile motion, r(t)=(x0+v0t)i+(h+v0yt12gt2)jCalculation: The given projectile motion is two-dimensional projectile motion.

The vector position of bomb is r1(t) and the position vector of the projectile is r2(t). The initial velocity of projectile is v at an angle of θ.

Horizontal component of velocity of projectile is,

vox=vcosθ

The vertical component of velocity of projectile is,

voy=vsinθ

The position vector of the bomb is written as,

r1(t)=(x0+v0t)i+(h+v0yt12gt2)j

Now, substituting h=3200, vox=400, voy=0, x0=5000 and g=32 in the above equation,

r1(t)=(5000400t)i+(3200+(0)t12(32)t2)j=(5000400t)i+(320016t2)j

The position vector of the projectile is written as,

r2(t)=(x0+v0t)i+(h+v0yt12gt2)j

Now, substituting h=0, vox=vcosθ, voy=vsinθ, x0=0 and g=9

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