An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9∘ above the horizontal. part a determine x-values at each 1 s from t = 0 s to t = 6 s determine vx- values at each 1s from t=0s to t=6s

An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9⁰.

Horizontal component of velocity = 50 cos 36.9 = 39.98 m/s

The horizontal velocity in projectile motion is constant.

So value from t = 0 seconds to t = 6 seconds is a constant, which is equal to 39.98 m/s.

a) Displacement along X-axis

    Displacement, X =  *t

     X (t=0) = 39.98*0 = 0 m

     X (t=1) = 39.98*1 = 39.98 m

     X (t=2) = 39.98*2 = 79.96 m

     X (t=3) = 39.98*3 = 119.94 m

     X (t=4) = 39.98*4 = 159.92 m

     X (t=5) = 39.98*5 = 199.90 m

     X (t=6) = 39.98*6 = 239.88 m

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.