A projectile of mass 0.552 kg is shot from a cannon, at height 7 m,?

In summary, a projectile with a mass of 0.552 kg is launched from a cannon at a height of 7 m with an initial horizontal velocity of 5 ms. The projectile reaches a maximum height of Y and lands at a horizontal distance X from the end of the cannon's barrel. The acceleration of gravity is 9.8 ms^2. To determine the vertical component of the initial velocity, the maximum height achieved, the magnitude of the velocity at impact, and the angle at which the projectile hits the ground, some work is required before obtaining the answers. The range of the projectile can also be calculated and the answers will be in units of ms and m.
  • #1
ahsanmhamid
9
0
A projectile of mass 0.552 kg is shot from a
cannon, at height 7 m,
with an initial velocity vi having a horizontal
component of 5 ms.
The projectile rises to a maximum height of
Y above the end of the cannon's barrel and
strikes the ground a horizontal distance X
past the end of the cannon's barrel.
The acceleration of gravity is 9.8 ms^2 :

Determine the vertical component of the
initial velocity at the end of the cannon's bar-
rel, where the projectile begins its trajectory.
Answer in units of ms.

Determine the maximum height Y the pro-
jectile achieves after leaving the end of the
cannon's barrel. Answer in units of m.

Find the magnitude of the velocity vector
when the projectile hits the ground. Answer
in units of ms.

Find the magnitude of the angle (with repect
to horizontal) the projectile makes when im-
pacting the ground. Answer in units of degree.

Find the range"x" of the projectile. Answer
in units of m.
 
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  • #2
Again, show some work before we can help you.
 
  • #3
i need some help setting up the problem.
 

1. What is the initial velocity of the projectile?

The initial velocity of the projectile can be calculated using the formula v = √(2gh), where v is the initial velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the initial height (7 m). Substituting these values, we get v = √(2*9.8*7) = 14.9 m/s.

2. How far will the projectile travel horizontally?

The horizontal distance traveled by the projectile can be calculated using the formula d = v*t, where d is the distance, v is the horizontal velocity, and t is the time. Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the projectile's flight. Therefore, the distance traveled horizontally will depend on the time the projectile is in the air. To calculate the time, we use the formula t = √(2h/g), where h is the initial height. Substituting the values, we get t = √(2*7/9.8) = 1.38 seconds. Therefore, the projectile will travel a horizontal distance of d = 14.9*1.38 = 20.6 meters.

3. What is the maximum height reached by the projectile?

The maximum height reached by the projectile can be calculated using the formula h = (v^2)/(2g), where h is the maximum height and v is the initial velocity. Substituting the values, we get h = (14.9^2)/(2*9.8) = 11.2 meters.

4. How long will the projectile be in the air?

The time the projectile is in the air can be calculated using the formula t = √(2h/g), where h is the initial height. Substituting the values, we get t = √(2*7/9.8) = 1.38 seconds.

5. What is the maximum height reached by the projectile?

The maximum height reached by the projectile can be calculated using the formula h = (v^2)/(2g), where h is the maximum height and v is the initial velocity. Substituting the values, we get h = (14.9^2)/(2*9.8) = 11.2 meters.

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