Answer :
Complete question
A cannon with a muzzle velocity of 500. meters per second fires a cannonball at an angle of 30.° above the horizontal. What is the vertical component of the cannonball's velocity as it leaves the cannon?
A 0.0 m/s
B 250. m/s
C 433 m/s
D 500. m/s
Answer:
The correct option is C
Explanation:
From the question we are told that
The velocity is [tex]v = 500 \ m/s[/tex]
The angle is [tex]\theta = 30^o[/tex]
Generally the vertical component of the canon ball is mathematically represented as
[tex]v_y = v * cos (30 )[/tex]
=> [tex]v_y = 500 * cos (30 )[/tex]
=> [tex]v_y = 433 \ m/s[/tex]
Answer:
250. m/s
Explanation:
It is vertical component so you do: 500mSin(30°) = 250m/s