Thank you for visiting A projectile is fired with a velocity of 45 m s at an angle of 32 What is the horizontal component of the velocity A. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
The horizontal component of velocity of the projectile, which is fired velocity with 45 m/s at an of 32° is 38.2 m/s.
What is projectile motion?
Projectile motion is the motion of the body, when it is thrown in the air taking the action of gravity on it.
Given information-
The initial velocity of the projectile is 45 meter per second.
The initial angle of fire is 32 degrees.
As the initial velocity of the projectile is 45 meter per second and the initial angle of fire is 32 degrees.
Thus, the horizontal component of velocity can be represent with the right angle triangle shown in the attached image below.
Suppose the horizontal component of the velocity is [tex]v_x[/tex]. Thus the cosine angle from the figure can be given as,
[tex]\cos(32)=\dfrac{v_x}{32}\\v_x=32\times \cos(32)\\v_x=38.2\rm m/s[/tex]
Thus the horizontal component of the velocity is 38.2 meter per second.
Learn more about the projectile motion here;
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Rewritten by : Jeany
Answer:
C
Explanation:
To calculate adjacent of triangle use
[tex] \cos(32) = \frac{a}{45} [/tex]
where 45 is the hypotenuse, and a is the adjacent side (horizontal component)
[tex]45 \cos(32) = 38.16[/tex]
then round to 38.2
