Question

An artillery shell is launched on a flat, horizontal field at an angle of α = 31.7° with respect to the horizontal and with an initial speed of v0 = 202 m/s. What is the horizontal velocity of the shell after 18.96 s of flight? (Neglect air friction. Use the coordinate system where the x-axis is horizontal and points to the right; and the y-axis is vertical and points up.)

Answer 1

171.86 m/s

Step-by-step explanation:

vo = 202 m/s

α = 31.7°

t = 18.96 s

In case of projectile motion, the acceleration along the horizontal direction is zero. It means the magnitude of velocity along horizontal direction is always constant. The horizontal component of velocity is given by

vx = vo Cosα

vx = 202 Cos 31.7

vx = 171.86 m/s

Thus, the horizontal component of velocity is 171.86 m/s.

Answer 2

Answer:193.90 m/s

Step-by-step explanation:

Given

launch angle
`\theta =31.7^(\circ)`

launch velocity
`v_0=202 m/s`

Horizontal velocity of the shell after
`t=18.96 s`

time of flight of Projectile
`T=(2u\sin \theta )/(g)`

`T={\frac2* 202\sin 31.7}{9.8}`

`T=21.66 s`

i.e. projectile is declining as
`t>(T)/(2)`

but horizontal component of velocity will remain same as there is no opposing force in horizontal direction

Horizontal component of velocity is

`u_x=v_0\cos \theta =202\cdot \cos 31.7=193.90 m/s`