Ajet plane is flying at a constant altitude. at time t1=0 it has components of velocity vx= 94 m/s , vy= 110 m/s . at time t2= 33.5 s the components are vx= 175 m/s , vy= 45 m/s . part a for this time interval calculate the average acceleration. give your answer as a pair of components separated by a comma. for example, if you think the x component is 3 and the y component is 4, then you should enter 3,4. express your answer using two significant figures.

The motion where it is is acted upon two forces of different directions is a characteristic of a projectile motion. These two vector forces of the resultant force are the vx and vy component forces. The resultant force is denoted by v only, while the components are vx and vy. 

Now, the are to determine the x and y components of acceleration also denoted as ax and ay. By definition, acceleration is the change of velocity per change of time:

a = Δv/Δt

If we distinguish the component vectors,

ax = Δvx/Δt    and ay = Δvy/Δt

The value for Δt is: Δt = 33.5 - 0 = 33.5 s. So, substituting the values,

ax = (vx₂ - vx₁)/Δt  
ax = (175 m/s - 94 m/s)/33.5
ax = 2.42 m/s²

ay = (vy₂ - vy₁)/Δt  
ay= (45 m/s - 110 m/s)/33.5
ay = -1.94 m/s²

Hence, the component vectors of the acceleration in (2.42,-1.94).

Fitness is the ability to perform only very light levels of physical activity.

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward