Question

A parachutist descending at a speed of 10ms drops a camera at an altitude of 50m . How long does it take the camera to reach the ground . What is the velocity of the camera just before it hits the ground.​

Answer 1

Given : -

A parachutist descending at a speed of 10 m/s drops a camera at an altitude of 50 meters .

Required to find : -

• Time taken by the camera to reach the ground ?
• Velocity of the camera with which it hits the ground ?

Equations used : -

• s = ut+½gt²
• v²-u²=2as

Here,

s = displacement

u = initial velocity

v = final velocity

g = acceleration due to gravity

t = time taken

Solution : -

A parachutist descending at a speed of 10 m/s drops a camera at an altitude of 50 meters .

From this given information we can conclude that;

Initial velocity of the parachutist is 10 m/s

Since,

He is the one who drops the camera .So, the initial velocity of the camera (u) is also 10 m/s

Altitude from which it was dropped is 50 meters. so, displacement (s) = 50 meters

The body is in freefall it experiences acceleration due to gravity . so, g = 10 m/s² ( approximately )

Using the 2nd Equation of motion i.e.

s = ut+½at²

50 = (10)(t)+½(10)(t)(t)

50 = 10t+(5)(t²)

50 = 10t+5t²

0 = 10t+5t²-50

10t+5t²-50=0

5t²+10t-50=0

5(t²+2t-10)=0

t²+2t-10=0/5

t²+2t-10=0

Now,

Let's find the roots of the quadratic equation !

Quadratic formula;

x = (-b±√[b²-4ac])/(2a)

Here,

• x = t
• a = 1
• b = 2
• c = -10

t = (-2±√[{2}²-4{1}{-10}])/(2{1})

t = (-2±√[4+40])/(2)

t = (-2±√[44])/(2)

t = (-2±√[4 x 11])/(2)

t = (-2±2√[11])/(2)

t = (-2+2√[11])/(2) & (-2-2√[11])/(2)

t = (-2+2 x 3.316)/(2) & (-2-2 x 3.316)/(2)

t = (-2+6.632)/(2) & (-2-6.632)/(2)

t = (4.632)/(2) & (-8.632)/(2)

t = 2.316 & -4.316

since,

Time can't be in negative

t = 2.316 seconds

Hence,

Time taken by the camera to reach the ground (t) = 2.316 seconds

Now,

Let's find the final velocity of the camera .

Using the 3rd Equation of motion i.e.

v² - u² = 2as

v² - (10)² = 2(10)(50)

v² - 100 = 20 x 50

v² - 100 = 1000

v² = 1000 + 100

v² = 1100

v = √(1100)

v = ±√(25 x 44)

v = ±5√(44)

v = ±5√(4 x 11)

v = ±5 x 2√(11)

v = ±10√(11)

v = ±10 x 3.316

v = ±33.16

since, velocity had no chance of being negative .

So,

v = 33.16 m/s

Hence,

Velocity with which the camera hits the ground is 33.16 m/s .

Answer 2

Answer:

• Distance (s) = 50 m
• Initial Velocity (u) = 10 m/s
• Acceleration (g) = 10 m/s²

• By Second Motion of Gravitation :

$:\implies\sf s = ut+\dfrac{1}{2}gt^2\\\\\\:\implies\sf50 = 10t + \dfrac{1}{2} \times 10 \times t^2\\\\\\:\implies\sf50 = 10t + 5t^2\\\\\\:\implies\sf10 = 2t + t^2\\\\\\:\implies\sf t^2 + 2t - 10 = 0\\\\\\:\implies\sf t = \dfrac{ -2 \pm\sqrt{(2)^2 -4 \times 1 \times - 10}}{2 \times 1}\\\\\\:\implies\sf t = \dfrac{ - 2 \pm \sqrt{44} }{2}\\\\\\:\implies\sf t = \frac{2( - 1 \pm \sqrt{11})}{2}\\\\\\:\implies\sf t = - 1 \pm3.31\\\\\\:\implies\sf t = 2.31 \:s\: \:or \: \:t = - 4.31 \:s$

$\rule{180}{1.5}$

• Using Positive value of t in Equation 1 :

$:\implies\sf v = u + gt\\\\\\:\implies\sf v = 10+10(2.31)\\\\\\:\implies\sf v=10+23.1\\\\\\:\implies\underline{\boxed{\sf v=33.1\:m/s}}$

⠀

$\therefore\:\underline{\textsf{Velocity of camera before it hits the ground is \textbf{33.1 m/s}}}.$