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A mortar shell is fired with a muzzle speed of 300 ft/sec. Find the angle of elevation of the mortar if the shell strikes a target located 1500 ft away. Round your answer to 2 decimal places.

Respuesta :

Answer:

Angle of elevation = 4.37°

Step-by-step explanation:

Velocity of the mortar shell = 300ft/sec

Distance covered= 1500ft

Range = U²sin2tita/g

But range= distance

Range = 1500ft

U = velocity

g = acceleration due to gravity

Tita = angle of projection of elevation

1500 = 300²sin(2tita)/9.87

(1500*9.87)/90000= sin2tita

0.1645 = sin2tita

Sin^-1 0.1645 = 2tita

9.4682 = 2tita

total = 9.4682/2

tita = 4.3741°

To two decimal place= 4.37°

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