   Chapter 3, Problem 39AP

Chapter
Section
Textbook Problem

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 100. m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 30.0 m/s2. At this time, its engines fail and the rocket proceeds to move as a projectile. Find (a) the maximum altitude reached by the rocket, (b) its total time of flight, and (c) its horizontal range.

(a)

To determine
The maximum altitude reached by the rocket.

Explanation

The distance travelled is found from the kinematic equation,

s=v0t+12at2

Here,

v0 is the initial velocity

a is the acceleration

t is the time taken

Substitute 100m/s for v0 , 3.00s for t and 30.0m/s2 for a .

s=(100m/s)(3.00s)+12(30.0m/s2)(3.00s)2=435m

The coordinates of the rocket at the end of the powered flight are,

x1=scosθy1=ssinθ

Substitute 435m for s and 53.0° for θ .

x1=(435m)cos53.0°=262my1=(435m)sin53.0°=347m

The speed of the rocket at the end of the powered flight is,

v1=v0+at

Substitute 100m/s for v0 , 3.00s for t and 30.0m/s2 for a .

v1=100m/s+(30.0m/s2)(3.00s)=190m/s

The components of the initial velocity of the rocket are,

v0x=v1cosθv0y=v1sinθ

Substitute 190m/s for v1 and 53.0° for θ .

v0x=(190m)cos53

(b)

To determine
The total time of flight.

(c)

To determine
The horizontal range.

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