Question

An artillery shell is launched on a flat, horizontal field at an angle of α = 42.6° with respect to the horizontal and with an initial speed of v₀ = 327 m/s.

a) What is the horizontal distance covered by the shell after 3.73 s of flight?

b) What is the height of the shell at this moment?

Answer 1

Gravitational acceleration = g = -9.81 m/s²

Initial velocity of the artillery shell = V₀ = 327 m/s

Angle of launch = $\alpha$ = 42.6^o

Initial horizontal velocity of the artillery shell = V_x0

V_x0 = V₀Cos$\alpha$

V_x0 = (327)Cos(42.6)

V_x0 = 240.704 m/s

Initial vertical velocity of the artillery shell = V_y0

V_y0 = V₀Sin$\alpha$

V_y0 = (327)Sin(42.6)

V_y0 = 221.338 m/s

Time period = T = 3.73 s

There is no horizontal force acting on the shell hence the velocity of the shell in the horizontal direction doesn't change.

Horizontal distance covered by the shell in 3.73 sec = R

R = V_x0T

R = (240.704)(3.73)

R = 897.83 m

Height of the shell at 3.73 sec = H

H = V_y0T + gT²/2

H = (221.338)(3.73) + (-9.81)(3.73)²/2

H = 757.35 m

a) Horizontal distance covered by the shell after 3.73 sec of flight = 897.83 m

b) Height of the shell at this moment = 757.35 m