A plane, diving with constant speed at an angle of 53.0 degrees....

In summary, a plane diving at a constant speed and angle of 53 degrees with the vertical released a projectile at an altitude of 730m. The projectile hit the ground after 5.00s. To find the speed of the plane, we can use the equation v = d/t where d is the vertical displacement and t is the time. To find the horizontal distance traveled by the projectile, we can use the equation d = v0t + 1/2at^2 where v0 is the initial velocity, a is acceleration, and t is the time. The horizontal component of the projectile's velocity just before striking the ground can be found using the equation v0x = v0cosθ, where v0 is
  • #1
HappyFlower
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Homework Statement


A plane, diving with constant speed at an angle of 53 degrees with the vertical, releases a projectile an altitude of 730m. The projectile hits the ground 5.00s after release. A plane, diving with constant speed at an angle of 53 degrees with the vertical, releases a projectile an altitude of 730m. The projectile hits the ground 5.00s after release. A) What is the speed of the plane? B) how far does the projectile travel horizontally during its flight? C) what are the horizontal and the vertical components of its velocity just before striking the ground?

Homework Equations

The Attempt at a Solution


I am having trouble understanding why my component will be -voxcos53* either than that I understand how to solve the problem.
 
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  • #2
HappyFlower said:
I am having trouble understanding why my component will be -voxcos53* either than that I understand how to solve the problem.
Which component is this? Usually components are written either as v0x or v0cosθ. I suspect you mean -v0cos53o. The negative sign is there because the plane is diving not ascending.
 
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  • #3
kuruman said:
Which component is this? Usually components are written either as v0x or v0cosθ. I suspect you mean -v0cos53o. The negative sign is there because the plane is diving not ascending.

I overthought that too much... thank you for the clarification.
 

FAQ: A plane, diving with constant speed at an angle of 53.0 degrees....

What is the acceleration of the plane?

The acceleration of the plane can be calculated using the formula a = g*sin(53.0), where a is the acceleration and g is the acceleration due to gravity (9.8 m/s^2). The answer is approximately 8.0 m/s^2.

What is the velocity of the plane?

The velocity of the plane can be calculated using the formula v = v0 + at, where v0 is the initial velocity (which is assumed to be 0 in this case), a is the acceleration, and t is the time. Since the plane is diving with constant speed, the velocity remains constant throughout the dive and is equal to the initial velocity. Therefore, the velocity of the plane is 0 m/s.

What is the angle of the plane's trajectory?

The angle of the plane's trajectory is given as 53.0 degrees in the question. This angle is measured from the horizontal plane and is also known as the angle of inclination or the angle of elevation.

What is the distance traveled by the plane during the dive?

The distance traveled by the plane can be calculated using the formula d = v0*t + (1/2)*a*t^2, where d is the distance, v0 is the initial velocity, a is the acceleration, and t is the time. Since the plane is diving with constant speed, the initial velocity is 0 and the acceleration is constant. Therefore, the distance traveled by the plane is equal to (1/2)*a*t^2, which is approximately 16.0 m.

What is the final velocity of the plane before it hits the ground?

The final velocity of the plane can be calculated using the formula v = v0 + at, where v0 is the initial velocity, a is the acceleration, and t is the time. Since the plane is diving with constant speed, the acceleration is constant and the initial velocity is 0. Therefore, the final velocity of the plane before it hits the ground is equal to the acceleration multiplied by the time, which is approximately 8.0 m/s.

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