Two-Dimensional Kinematics, Projectile

[Naeem]

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

 

[whozum]

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

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

$$t = \sqrt{\frac{2d}{g}}$$

 

[thepatientmental]

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.