A 2.00-kilogram mud ball drops from rest at a height of 17.0 m. If the impact between the ball and the ground lasts 0.46 s, what is the magnitude of the average force exerted by the ball on the ground

79.35 N

Explanation: according to the impulse momentum theorem,

Impulse = change in momentum

Where impulse = force × time and change in momentum = m ( v - u).

The object was initially at rest, hence it initial velocity is zero.

To get the final velocity, we use the formulae below

v² = u² + 2gh

Where h = height of the cliff = 17m

v² = 2 × 9.8 × 17

v² = 333.2

v = √333.2

v = 18.25 m/s

At t = 0.46s and v = 18.25 m/s, we can get the average force of impact

F×0.46 = 2 (18.25 - 0)

F × 0.46 = 2 (18.25)

F × 0.46 = 36.5

F = 36.5 /0.46 = 79.35 N

diagram is missing, but i will try to explain your doubt.

answer:

amplitude is the maximum displacement or distance moved on a point of a vibrating body or wave measured from its equilibrium position.it is equal to half the length of vibration path.

ill do it for 100, also this isnt 80 its 40