   Chapter 13.4, Problem 24E

Chapter
Section
Textbook Problem

# Rework Exercise 23 if the projectile is fired from a position 100 m above the ground.

(a)

To determine

To find: The range of the projectile.

Explanation

Given:

v0=200ms , θ=60° and r(0)=100j .

Formula:

Write the expression for second law of motion.

F(t)=ma(t) (1)

Here,

F(t) is the force,

m is the mass, and

a(t) is the acceleration.

Write the expression to find the acceleration.

a(t)=v(t) (2)

Write the expression to find the velocity.

v(t)=r(t) (3)

Consider the unit vector in the direction of the velocity.

v(0)=v0[cos(θ)i+sin(θ)j]

Substitute 60° for θ and 200ms for v0 ,

v(0)=200[cos(60°)i+sin(60°)j]=200[12i+32j]=100i+1003j

Write the expression of force due to gravity, by ignoring the air resistance.

F(t)=mgj (4)

Here,

g is the gravity, which is 9.8ms2 .

Equate equation (1) and equation (4).

ma(t)=mgja(t)=gj

Substitute 9.8ms2 for g ,

a(t)=9.8j

Modify equation (2).

v(t)=a(t)dt

Substitute 9.8j for a(t) ,

v(t)=9.8jdt

v(t)=9.8tj+C (5)

Here v(0)=C .

Substitute 100i+1003j for v(0) ,

C=100i+1003j

Substitute 100i+1003j for C in equation (5),

v(t)=9.8tj+100i+1003j=100i+(10039.8t)j

Modify equation (3).

r(t)=v(t)dt

Substitute 100i+(10039

(b)

To determine

To find: The maximum height reached.

(c)

To determine

To find: The speed at impact.

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