[SOLVED] What is the ball's impact speed on the ground below?

Question

What is the ball's impact speed on the ground below?

A cannon tilted up at a 31.0 angle fires a cannon ball at 66.0m/s from atop a 25.0m -high fortress wall.

What is the ball's impact speed on the ground below?

Answer 1

The Best answer for What is the ball's impact speed on the ground below?

Using Conservation of Energy:

Ek + Eg = Ek

1/2 m vi^2 + mgh = 1/2 m vf^2

vf = sqrt (vi^2 + 2gh)

vf = sqrt (66^2 + 19.6(25))

vf = 69.6 m/s

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.

SHOW ANSWER

Answer 2

In the two situations, the preliminary complete ability of the ball is m*h + ½*m*v²; so the kinetic ability whilst the ball hits the floor is an identical, and the speed is an identical. to seem at it yet in a distinctive way, if the ball is thrown up with speed v, it ought to have this comparable speed interior the downward direction whilst it passes the altitude of the cliff. for this reason the tip result's comparable to if the ball replaced into thrown downward with that speed.

i think this is the correct answer for What is the ball's impact speed on the ground below?

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.

SHOW ANSWER

Answer 3

The initial kinetic energy is:

KEi = (1/2)m(vi)^2

The final kinetic energy is:

KEf = (1/2)m(vf)^2

The initial potential energy is:

PEi = mgh

The final potential energy is:

PEf = mg(0) = 0

Since energy is conserved:

KEf + PEf = KEi + PEi

(1/2)m(vf)^2 + 0 = (1/2)m(vi)^2 + mgh

Multiply both sides by 2 and divide both sides by m:

vf^2 = vi^2 + 2gh

Take the square root of both sides, and plug in given values:

vf = sqrt[vi^2 + 2gh] = sqrt[(66.0 m/s)^2 + 2(9.81 m/s^2)(25.0 m)] = 69.6 m/s

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.

SHOW ANSWER