Ashell if fired from the ground with an initial velocity of 1,700 m/s at an initial angle of 55 degrees to the horizontal. neglecting air resistance, what is the shell's horizontal range?

Therefore the horizontal range = 294897.96 m.

Explanation:

Range of a projectile: The range is defined as the horizontal distance from the point of projection to the point where the projectile hit the projection plane again. The S.I unit of range is Meter (m).

It can be expressed mathematically as

R = u²sin2∅/g Equation 1

Where R = Horizontal range, ∅ = angle of projection, u = initial velocity, g = acceleration due to gravity.

Given: u = 1700 m/s, ∅ = 55°,

Constant: g = 9.8 m/s²

Substituting these values into equation 1

R = (1700²sin55)/9.8

R = 2890000/9.8

R = 294897.96 m.

Therefore the horizontal range = 294897.96 m.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s