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A stone is thrown vertically upward with a speed of 12.5 m/s from the edge of a cliff 75.0 m high

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Answer #1
Concepts and reason

The concepts required to solve this question are the kinematic equations.

The physical quantities such as distance travelled, time taken for motion, initial and final velocity of a moving object are related to each other by kinematic equation of motion.

If an object is thrown vertically upwards, the displacement of the object in travelling up and back to its initial position is zero.

Fundamentals

Consider an object thrown upwards with initial speed u from a cliff of depth below origin. Let the initial position of the object be y distance above the origin.

The time t taken by the object for its complete motion is given by following expression:

glº+ut + y = y

The final velocity v of the object returning towards ground under acceleration a is given by following expression:

v=u + at

Let Vi
be the velocity of the object at the highest point of motion and its height above origin be h and given by following expression:

v* = u? +2ah

The total distance s travelled by the object is the sum of distance travelled in going up to highest point, returning up to initial point and the vertical distance of cliff form origin.

s=yo+h+h
.

[Starting Hint]

Use kinematic equation of motion for calculating time taken for motion.

The distance travelled by an object can be calculated by using kinematics equation of motion and given below:

y=yo +ut, +3 at

Here, is the time taken by the object for its in travelling from top of cliff to the bottom.

Substitute (0 m)
for y, (-75.0 m)
for , (9.8 m·s)
for a and (14.0 m-s)
for u in the last expression as follows:

(0 m)=(-75.0 m)+(14.0 m-s
$?)

Rearrange this expression for time as follows:

(-75.0 m)+(14.0 m-s-)+(9.8 m-s2){ = 0 m
4.913 +14.01, - 75=0

Solve this quadratic equation as follows:

14.9 +14.02.-75 =0
(1 +5.592)(1, -2.736)=0
t = -5.59,2.736

As time is always taken positive, so the negative value is discarded.

So, the total time t is given as follows:

t=1, +1, +12

Substitute (2.736 s)
for and (1.428 s)
for as follows:

1 = 1, +1, +12
= (1.428 s)+(1.428 s)+(2.736 s)
= 5.592 s

(b)

Find the speed of the object before hitting the ground.

The speed of the object at the level of cliff, while returning is same as that of the initial velocity.

So, the final velocity v with which, the object hits the ground is given as follows:

v=u+at,

Substitute (2.736s)
for , (9.8 m·s)
for a and for u in the last expression as follows:

v=u+at
= (14.0 m-s*)+(9.8 m-s?)(2.736 s)
= 40.81 m.s?

(c)

Find the distance travelled by the object before hitting the ground.

v} = u’ +2ah
h=
2a

Substitute (0 m)
for Vi
,(-9.8 m·s)
for a and for u in the last expression as follows:

2a
(0 m-s) - (14.0 m-s)?
* 2 (-9.8 m-s?)
= 10.0 m

Total distance travelled by the object is given as follows:

s=yo+h+h

Substitute (10.0 m)
for h and (75.0 m)
for as follows:

s = y; +h+h
= (75.0 m)+(10.0 m)+(10.0 m)
= 95.0 m

Ans: Part a

The object reach the bottom of the cliff after (5.59s)
.

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