A projectile is fired horizontally from a gun that is 58.0 m above flat ground, emerging from the gun with a speed of 170 m/s. (a) How long does the projectile remain in the air

t = 3.44 s

Explanation:

We are given;

Fired from rest, and so; u = 0 m/s

Final speed; v = 170 m/s

Height above flat ground; y_o = 58 m

Height at starting point; y = 0 m

Thus, from Newton's equation of motion, we have;

y - y_o = ut - ½gt²

(since it's motion is against gravity)

Plugging in the relevant values, we have;

0 - 58 = 0 - (½ × 9.8 × t²)

-58 = -4.9t²

t² = 58/4.9

t = √(58/4.9)

t = 3.44 s

wavelegth as in spectrum red oygbiv

98.4 m

explanation:

let s = distance up the plane  

then by conservation of energy  

1/2m(18.3)^2 = mg s sin(10)  

(18.3)^2 = 2g s sin(10)  

s = (18.3)^2/2gsin(10)  

= 98.39564526  

= 98.4 m