Gun was fired with a muzzle velocity of 350m/s, mounted at an angle of 45’ above the ground. Neglecting air resistance, compute for the following; *Maximum height reached
*Range of the projectile
*Total time of flight

Maximum height, h = 3062.5m

Total time of flight, T = 49.49secs or 50secs approx.

Range, R = 12250m

Explanation:

Given data:

U = 350m/s

Angle = 45°

Assume g = 10m/s

At the greatest height, v = 0

Therefore,

V^2 = U^2 sin^2 × angle - 2×g×h

Substituting values:

0^2 = 350^2 sin^2 (45) - 2 × 10 × h

Let h = maximum height reached

Rearranging gives:

350^2 sin^2(45) = 2 x 10 x h

h = 350^2 sin^2(45)/2×10

h = 122500 x 0.5/20

h = 61250/20

h = 3062.5m

2)Total time of flight, T

T = 2U sin(angle)/g

= 2x350 sin(45)/10

= 494.9747/10

= 49.49secs or 50sec approx.

3) Range of projectile, R

R = U^2 sin2(angle)

= 350^2 sin2 (45)

= 122500 x 1/10

= 12250m

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form