Welcome to the article During a fireworks display a shell is shot into the air at an angle above the horizontal The fuse is timed so that the shell. On this page, you will learn the essential and logical steps to better understand the topic being discussed. We hope the information provided helps you gain valuable insights and is easy to follow. Let’s begin the discussion!
Answer :
The initial velocity angle is 49.4 degrees.
Using the projectile motion equation of motion, we can compute the beginning velocity of the shell as follows:
v = √(2gh)
Where h is the height of the shell when it explodes (51.6 meters) and g is the acceleration caused by gravity (9.8 m/s2).
Consequently, v = √(2 × 9.8 × 51.6) = 46.3 m/s
The angle of the starting velocity may then be determined using the projectile motion equation of motion:
tan(θ) = vy / vx
where vy denotes the velocity's vertical component and vx denotes its horizontal component.
The following equation may be used to compute the vertical component of the velocity:
vy = v × sin(θ)
where v is the starting speed (46.3 m/s) and is the starting speed's angle
The following equation may be used to get the horizontal component of the velocity:
vx = v × cos(θ)
where is the angle of the initial velocity and v is the initial velocity (46.3 m/s) (which we are trying to find).
As a result, tan(θ) = (46.3 × sin(θ)) / (46.3 × cos(θ))
By solving for θ, we obtain:
θ = tan-1(sin(θ) / cos(θ)) = tan-1(0.948) = 49.4 degrees
Therefore, the initial velocity's of angle θ is 49.4 degrees.
Complete Question:
During a fireworks display, a shell is shot into the air at an angle of θ degrees above the horizontal. The fuse is timed so that the shell explodes after 9.7 seconds, just as it reaches its highest point above the ground. If the horizontal displacement of the shell is 51.6 meters when it explodes,what is the angle θ of the initial velocity? Use g=9.8 m/s 2
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Rewritten by : Brahmana