Two-Dimensional Kinematics, Projectile

In summary, at t = 10 s into the flight, the vertical acceleration of the object is -9.81 m/s2, its horizontal acceleration is 0 m/s2, its vertical velocity is -83.1 m/s, its horizontal velocity is 25.98 m/s, the angle at which it is traveling is -72.59 degrees, its vertical displacement is -340.5 m, and its horizontal displacement is 259.8 m. To find the time at which the object reaches its maximum height, you can use the equation t = √(2d/g), where d represents the vertical displacement.
  • #1
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Q. At t = 0, an object is projected with a speed v0 = 30 m/s at an angle q0 = 30° above the horizontal. The axes on the diagram show the x and y directions that are to be considered positive.

For parts a-g, calculate the requested quantities at t = 10 s into the flight. ( Use 9.81 m/sec2 for g. )


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a) The vertical acceleration of the object:
ay = m/s2 *

HELP: What force is acting on the object in the vertical direction?
HELP: Is it moving faster or slower as it travels upward?


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b) Its horizontal acceleration:
ax = m/s2 *
0 OK


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c) Its vertical velocity:
vy = m/s *
-83.1 OK

HELP: Use the definition of acceleration.


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d) Its horizontal velocity:
vx = m/s *
25.98 OK


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e) The angle to the horizontal at which the object is traveling (an angle above the horizontal should be reported as a positive number; an angle below the horizontal should be reported as a negative number). Please give your answer in degrees:
angle = ° *
-72.59 OK


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f) Its vertical displacement, from where it started:
y = m *
-340.5 OK

HELP: Look at the equations in your book.
HELP: Look for an equation relating initial velocity, change in position, acceleration, and time. Use your previous answers.


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g) Its horizontal displacement, from where it started:
x = m *
259.8 OK


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h) At what time does the object reach its maximum height?
ty, max = s
2*83.1*sin(72.59)/9.81 NO


I just need some help with part h.

At the max, height final velocity is zero, I know that

I tried using t = v_y/g

and plugging this into

y-yo = v0sintheta*t - 1/2 g* t^2

didn't work!

Pl. Help
 

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  • #2
I tried using t = v_y/g isn't right.

For an object under constant acceleration with no initial velocity it takes

[tex] t = \sqrt{\frac{2d}{g}}[/tex]
 
  • #3


To find the time at which the object reaches its maximum height, we can use the equation y-y0 = v0y*t - 1/2*g*t^2, where y0 is the initial vertical position, v0y is the initial vertical velocity, and g is the acceleration due to gravity. We know that at the maximum height, the vertical velocity is zero, so we can set v0y*t = 0 and solve for t. This gives us t = v0y/g. Plugging in the values from part c, we get t = (30*sin(30))/9.81 = 1.53 seconds. Therefore, the object reaches its maximum height at 1.53 seconds into the flight.
 

What is two-dimensional kinematics?

Two-dimensional kinematics is the study of the motion of objects in two dimensions, typically represented by x and y coordinates. This includes the analysis of an object's displacement, velocity, and acceleration in both the horizontal and vertical directions.

What is a projectile?

A projectile is any object that is launched into the air and moves along a curved path under the influence of gravity. Examples of projectiles include a thrown baseball, a launched rocket, or a kicked soccer ball.

What factors affect the trajectory of a projectile?

The trajectory of a projectile is affected by various factors, including its initial velocity, launch angle, air resistance, and the force of gravity. These factors determine the path that the object will follow as it moves through the air.

How do you calculate the range of a projectile?

The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the formula R = v0^2 * sin(2θ) / g, where v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

What is the difference between a projectile's range and its maximum height?

The range of a projectile is the horizontal distance it travels, while its maximum height is the highest point it reaches along its trajectory. These values are dependent on the launch angle and initial velocity of the projectile.

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