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A cannon tilted up at a 30 angle fires a cannon ball at 80 m/s from atop a 10-m-high fortress wall. what is the ball's impact speed on the ground below?

Respuesta :

carlosego
To find the velocity of the bullet just before reaching the ground, we use the formula.
 Vy = Vosen (30 °) + gt (1).
 For that we need to know the time it takes for the bullet to reach the ground from the moment it is fired.
 For that we use the formula:
 Xy = Xo + Vosen (30 °) * t + 0.5gt ^ 2 (2)
 where:

 Vy = Component of the speed in the y direction (vertical)
 Vo = initial velocity of the bullet at the exit of the cannon
 g = acceleration of gravity
 Xy = vertical component of the position.
 Xo = initial position.
 t = time.

 To find the time it takes for the bullet to reach the ground, we make Xy = 0 and clear t.
 then it would be:
 0 = 10 + 80sen (30 °) * t +0.5 * (- 9.8) t ^ 2.
 The solution to that equation is:

 t = 8.4 seconds.

 We substitute that time in equation (1) and clear Vy.
 We have left:

 Vy = 80 * sin (30 °) + 9.8 * (8.4)
 Vy = 42.38 m / s
onyebuchinnaji

The final velocity of the cannon ball is 82.35 m/s.

Time of motion of the cannon ball

The time of motion of the cannon ball is calculated as follows;

h = v₀t + ¹/₂gt²

10 = (80 x sin30)t + (0.5 x 9.8)t²

10 = 40t + 4.9t²

4.9t² + 40t - 10 = 0

solve the quadratic equation using formula method

t = 0.24 s

Final velocity of the cannon ball

The final velocity of the cannon ball is calculated as follows;

v = v + gt

v = 80 + 0.24(9.8)

v = 82.35 m/s

Learn more about final velocity of projectile here: https://brainly.com/question/82566

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