Answer:
v = 84.28 m/s
Explanation:
It is given that,
Initial velocity of a cannon, u = 83 m/s
Angle with which it is fired, [tex]\theta=29^{\circ}[/tex]
It is fired from a top a 11 m high fortress wall.
Let v is the ball's impact speed on the ground below. It can be calculated using the third equation motion.
[tex]v^2-u^2=2as\\\\v^2=2as+u^2\\\\v^2=2\times 9.8\times 11+(83)^2\\\\v=84.28\ m/s[/tex]
So, the ball's impact speed on the ground below is 84.28 m/s.
The ball's impact speed on the ground will be:
84.28 m/s
According to the question,
Cannon's initial velocity, u = 83 m/s
Angle, θ = 29°
Height, s = 11 m
By using third equation of motion,
→ v² - u² = 2as
or,
→ v² = 2as + u²
By substituting the values, we get
= 2 × 9.8 × 11 + (83)²
= 32.78 + 6889
= 6921.78
v = √6921.78
= 84.28 m/s
Thus the above answer is correct.
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