Question

Answer the following questions

Answer 1

Given:

Depth of the fall = 48m

Value of g= 9.8 m/s²

Initial Velocity = 0m/s

To Find:

A) How long did Munday fall to reach the water surface below the falls?

B) What was Munday's velocity as he reached the water surface?

C) What was Munday's velocity after a period of 1s?

D) When do we say that an object undergoes retardation?

Solution:

A) Since acceleration is constant, we can use 2nd equation of motion.

s = ut + 1/2 a t²

Since u = 0 m/s

and s = 48m

-48 = 1/2 (-9.8) t²

or t = 3.12 s

B) Using 1st equation of motion,

v = u + at

Substituting the value of t from (A)

v = 0 + (-9.8) (3.12)

v = - 30.57m/s

C) To find velocity after 1 s, we can use the 1st equation of motion.

v1 = u + at

taking t = 1 s

v1 = 0 + (-9.8) (1)

v1 = -9.8m/s

D) An object undergoes retardation when its net negative acceleration is in opposite direction to that of the object's velocity.

Answer 2

Answer:

Time =3.12s

v =30.57m/s

velocity after 1 s =9.8m/s

Explanation:

Initial velocity is =0m/s

Acceleration g= 9.8m/s

Depth of fall =48m

Time taken Munday fall to reach the water surface below the fall.

$s=ut+1/2at^{2}$

u =0m/s

s =48m

48 =1/2(9.8)2

t=3.12s

Munday 's velocity as he reach the water surface

$v=u+at$

v=0+9.8*3.12

v= 30.57m/s

Munday's velocity after a period of 1s

v'=u+ at

v' =o+(9.8)1

v'= 9.8m/s

We say the object goes retardation when negative force of speed oppose the velocity.

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