At t = 0, an object is projected with a speed v0 = 40 m/s at an angle q0 = 32.8o

Question

At t = 0, an object is projected with a speed v0 = 40 m/s at an angle q0 = 32.8o above the horizontal. The axes on the diagram show the x and y directions that are to be considered positive. For parts 1-7, calculate the requested quantities at t = 8 seconds into the flight.

1)The vertical acceleration of the object:

ay =

2)Its horizontal acceleration:

ax =

3)Its vertical velocity:

vy =

4)Its horizontal velocity:

vx =

5)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 =

6)Its vertical displacement, from where it started:

y =

7)Its horizontal displacement, from where it started:

x =

8)At what time does the object reach its maximum height?

ty, max =

Explanation / Answer

6.) use y=sy = uyt - 0.5gt2

  uy is 40 sin(32.8), t=8, g = 9.81m/s2

7.) x = uxt

ux=40cos(32.8) and t = 8

8.) vy=uy - gt = 0 where  uy is 40 sin(32.8), g = 9.81m/s2 calculate t

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