Question

A mortar* crew is positioned near the top of a steep hill. Enemy forces are charging up the hill and it is necessary for the crew to spring into action. Angling the mortar at an angle of θ = 57.0o (as shown), the crew fires the shell at a muzzle velocity of 241 feet per second. How far down the hill does the shell strike if the hill subtends an angle φ = 36.0o from the horizontal? (Ignore air friction.)

How long will the mortar shell remain in the air?

How fast will the shell be traveling when it hits the ground?

Answer 1

Given that :

Angling the mortar at an angle, = 57 degree

muzzle velocity, v = 241 ft/sec

(a) the shell strike if the hill subtends an angle = 36⁰ from the horizontal at a distance which is given as ::

on the vertical plane,   y = d Sin                                                        { eq.1 }

y = x Sin (-36⁰)

or   y = (-0.587) x                                                                                        { eq.2 }

using a trajectory equation :

y = h + x tan - g x² / 2 (v² . Cos²)                                                             { eq.3 }

where, h = 0 m

g = acceleration due to gravity = 32.2 ft/s²

inserting the values in eq.3,

(-0.587) x = (0 m) + x tan (57⁰) - (32.2 ft/s²) x² / 2 [(241 ft/sec)² Cos² (57⁰)]

-(0.587) x = (1.53) x - (32.2 ft/s²) x² / (116162 ft²/s²) (0.296)

-(0.587) x = (1.53) x - (32.2 ft/s²) x² / (34383.9)

-(0.587) x = (1.53) x - 0.00093 x²

-(0.587) x - (1.53) x = - 0.00093 x²

- (2.117) x = - 0.00093 x²

0.00093 x² - 2.117 x + 0 = 0                                       (quadratic equation)

x = 0 ft or x = 2276.3 ft

(b) time taken for the mortar shell remain in the air which will be given as :

on the horizontal plane,    from eq.2

x = (2276.3 ft) / Cos (-36⁰)

x = 2813.7 ft

and   y = (2813.7 ft) Sin (-36⁰)

y = (2813.7 ft) (-0.587)

y = -1651.6 ft

equation time of flight is given as :

t₁ = 2 v₀ Sin / g                                                                              { eq.4 }

inserting the values in eq.4,

t₁ = 2 (241 ft/sec) Sin 57⁰ / (32.2 ft/s²)

t₁ = (482 ft/s) (0.8386) / (32.2 ft/s²)

t₁ = 12.5 sec

initial vertical velocity is given as :

v_oy = v Sin                                                                                  { eq.5 }

inserting the values in eq.5,

v_oy = (241 ft/s) Sin 57⁰

v_oy = 201.1 ft/s

and returning initial vertical velocity to launch height, v_oy = -201.1 ft/s

Time taken to reach the ground which is given as :

using equation of motion 2,

y = v_oy t₂ - (1/2) g t₂²                                                                                 { eq.6 }

inserting the values in eq.6,

(-1651.6 ft) = (-201.1 ft/s) t₂ - (0.5) (32.2 ft/s²) t₂²

(-1651.6 ft) = (-201.1 ft/s) t₂ - (16.1 ft/s²) t₂²

(16.1) t₂² + (201.1) t₂ - (1651.6) = 0                                            (quadratic equation)

t₂ = -18.1 s or t₂ = 5.65 s

total time taken, T = t₁ + t₂ = (12.5 sec + 5.65 sec)

T = 18.1 sec

(c) the shell be traveling at velocity when it hits the ground which will given as ::

v = v_x² + v_y²                                                                         { eq.7 }

where, v_x = horizontal velocity = v cos

v_x = (241 ft/s) cos (57⁰) = 131.2 ft/s

and v_y = vertical velocity = v_oy - g t

v_y = (-201.1 ft/s) - (32.2 ft/s²) (5.65 s)

v_y = (-201.1 ft/s) - (181.9 ft/s) = -382.9 ft/s

inserting the value of v_x & v_y in eq.7,

v = (131.2 ft/s)² + (-382.9 ft/s)²

v = 163825.8 ft/s

v = 404.7 ft/s