Question

A student solving a physics problem for the range of a projectile has obtained the expression r= v20sin(2θ)g where v0=37.2meter/second, θ=14.1∘, and g=9.80meter/second2. use your calculator to evaluate r.

Answer 1

The range of a projectile is about 66.7 m

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Further explanation

Acceleration is rate of change of velocity.

$\boxed {a = \frac{v - u}{t} }$

$\boxed {d = \frac{v + u}{2}~t }$

where:

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

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Let's recall Range of Projectile formula as follows:

$\boxed{ R = \frac{ v_o^2 \sin 2\theta}{g} }$

where:

R = range of projectile ( m )

v₀ = initial speed of projectile ( m/s )

θ = angle of projection

g = gravitational acceleration ( m/s² )

Let us now tackle the problem!

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Given:

initial speed of projectile = v₀ = 37.2 m/s

angle of projection = θ = 14.1°

gravitational acceleration = g = 9.80 m/s²

Asked:

the range of a projectile = R = ?

Solution:

$R = \frac{ v_o^2 \sin 2\theta}{g}$

$R = \frac{ 37.2^2 \sin 2 (14.1^o)}{9.80}$

$\boxed {R = 66.7 \texttt{ m}}$

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The range of a projectile is about 66.7 m

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Learn more

• Velocity of Runner : brainly.com/question/3813437
• Kinetic Energy : brainly.com/question/692781
• Acceleration : brainly.com/question/2283922
• The Speed of Car : brainly.com/question/568302

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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Answer 2

Formula of range is given by

$R = \frac{v_o^2 sin2\theta}{g}$

here given that

$v_0=37.2m/s$

θ=14.1∘

$g=9.80m/s^2$

now by above equation

$R = \frac{37.2^2 sin2*14.1}{9.80}$

R = 66.7 m

so range will be 66.1 m