A cannon is fired 38 degrees above the horizontal with an initial speed of 58m/s

Question

A cannon is fired 38 degrees above the horizontal with an initial speed of 58m/s

a] what is the magnitude of the horizontal component of the cannon balls displacement at the end of 2 secs

b] how long does it take to reach its highest point

c] what is the acceleration 4 s after it is fired

d] what are the horizontal and vertical componants of the velocity Vx and Vfy at the end of 4s

e]what is the velocity at the end of 4s

Please be as detailed as possible

I know its alot but I would appriciate the help

Explanation / Answer

a)

Horizontal distance = Vx * t

= 58*cos(38 deg) * 2

=91.409 m

b)

time taken = Vy / g

t = 58*sin(38 deg) / 9.81

t = 3.64 s

c)

Since there is no acceleration in the horizontal direction, the velocity in the horizontal direction is constant. The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, being equal to g

a = 9.81 m/s^2

d)

Horizontal Vx = Vxo = 58*cos(38 deg) = 45.705 m/s

Vertical Vy = Vyo - gt = 58*sin(38 deg) - 9.81*4 = -3.5316 m/s

e)

Velocity = 45.705 i - 3.5316 j

Magnitude = sqrt(45.705^2 + 3.5316^2) = 45.84 m/s

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