Question

A 600 kg boulder falls from a cliff and hits the ground with a speed of 68.6 m/s. What is the height of the cliff?

Answer 1

$\huge {\underline{\pink{Given:-}}}$

• Mass of boulder ,m = 600kg
• Initial Velocity ,u = 68.6m/s
• Final velocity ,v = 0m/s
• Acceleration due to gravity ,a = 9.8m/s

$\huge {\underline{\red{To Find:-}}}$

• Height of cliff ,h

$\huge {\underline{\green{Solution:-}}}$

Using third Equation of Motion

• v² = u² +2ah

Substitute the value we get

➺ 0² = 68.6² + 2× (-9.8) ×h

➺ 0 = 4705.96 + (-19.6) ×h

➺ -4705.96 = -19.6×h

➺ h = -4705.96/-19.6

➺ h = 4705.96/19.6

➺ h = 240.1m

Therefore, the height of the cliff is 240.1 metres.

Answer 2

Explanation:

Given :

• A 600 kg boulder falls from a cliff .

• hits the ground with a speed of 68.6 m/s.

To Find :

• What is the height of the cliff?

Solution ;

_______________________

Concept :

Third equation of motion :

displaying constant acceleration. (Slope of the curve is constant) At t = 0 seconds, the particle’s velocity is u m/s.

At t = t seconds, the particle’s velocity is v m/s.

Area under the curve of v − t graph gives displacement.

Now,

S = Area of Rectangle + Area of Triangle

S = ut + 12(v - u)(t)

(Substituting t = (v - u)a, from first equation of motion)

S = u(v - ua) + 12(v - u)(v - ua)

Rearranging the terms,

v2 = u2 + 2aS

_______________________

We talked about writing height instead of distance

v² = u² +2ah

Substitute all values :

0² = 68.6² + 2 × (-9.8) × h

• ( (On multiple 2 and 9.8) )

- ( 68.6² ) = -19.6 × h

• ( on simplify the number )

- 4705.96 = - 19.6h

h = 4705.96/19.6

h = 240.1m

The height of the cliff is 240.1 metres.

more to know :

Velocity :

• The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of its speed and direction of motion.

Height

• Denote the maximum height obtained by a projectile launched with speed v at angle to the horizontal by . The height as a function of time t for a gravitational Acceleration g is given by. (1)

Acceleration :

• Acceleration is the rate of change of velocity of an object with respect to time. An object's acceleration is the net result of any and all forces acting on the object, as described by Newton's Second Law.