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

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

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