At t = 0, an object is projected with a speed v0 = 35 m/s at an angle q0 = 23.8o

Question

At t = 0, an object is projected with a speed v0 = 35 m/s at an angle q0 = 23.8o 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 = 6 seconds into the flight. (Use 9.81 m/sec2 for g.) I need answers for numbers 4-8

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

4) Vx= 35 cos 23.8 =32 m/s apprx

5)let's calculate the vertical velocity after 6seconds,

v = 35 sin 23.8 - 9.8 (6)=14.124 -58.8=-44.676 m/s

angle= tan^-1 (-44.676/32) = -54.387 degree apprx

6)y = ( 35 sin 23.8) (6) - 1/2 (9.8) 6^2=84.744-176.4= - 91.656 m apprx

7) x = 32 m/s apprx ( 6)= 192 m apprx

8)max height = ( 35 sin 23.8)^2/ 19.6= 10.2 m apprx

time to reach max heihht= 35 sin 23.8/9.8 =1.44 seconds apprx

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