A shell is fired from a cliff that is 30 m above a horizontal plane. The muzzle speed of the shell is 75.0 m/s and it is fired at an elevation of 35° above the horizontal.

(a) Determine the horizontal range of the shell.
Initial Velocity = 75 m/s
Elevation angle = 35°
Height of the cliff = 30 m

Range = (initial velocity) ^2 * sin(2θ)/g
Range = (75) ^2 * sin (2*35°) / 9.8 m/s
Range = 539.36 m

The horizontal range of the shell is approximately 539.36 m

(b) Determine the velocity of the shell as it strikes the ground.
Vertical Velocity = initial velocity * sin (θ)
Vertical velocity = 75 m/s * sin (35°)
Vertical Velocity = 42.8 m/s

The velocity of the shell is approximately 42.8 m/s

1 answer

as it strikes the ground.