Question

A hot air balloon has just lifted off and is rising at the constant rate of 1.60 m/s. suddenly, one of the passengers realizes she has left her camera on the ground. a friend picks it up and tosses it straight upward with an initial speed of 10.2 m/s. if the passenger is 2.60 m above her friend when the camera is tossed, how much time does it take for the camera to reach her?

Answer 1

It takes 0.388 seconds for the camera to reach her

Further explanation

These are the formulas that we have to remember before solving the problem.

Speed is the rate of change of distance.

$\large { \boxed {v = \frac{d}{t}}}$

v = speed ( m/s )

d = distance ( m )

t = time ( s )

Acceleration is the rate of change of velocity.

$\large { \boxed {a = \frac{\Delta v}{t}}}$

a = acceleration ( m/s² )

Δv = change in speed ( m/s )

t = time ( s )

Let us now tackle the problem!

Given:

Speed of Air Balloon = u = 1.60 m/s

Initial Speed of Camera = vo = 10.2 m/s

Initial Distance of Passenger and Camera = d = 2.60 m

Gravitational Acceleration = g = 9.80 m/s²

Unknown:

Time Required = t = ?

Solution:

When camera reaches her :

displacement of camera = 2.60 + displacement of passenger

$v_o ~ t - \frac{1}{2} ~ g ~ t^2 = 2.60 + u ~ t$

$10.2 ~ t - \frac{1}{2} ~ 9.80 ~ t^2 = 2.60 + 1.60 ~ t$

$4.9 ~ t^2 - 8.60 ~ t + 2.60 = 0$

To solve this quadratic equation, we can use the following formula:

$t= \frac{-b - \sqrt{b^2-4ac}}{2a}$

$t = \frac{8.60 - \sqrt{(-8.60)^2 - 4(4.9)(2.60)} }{2(4.9)}$

$\boxed {t \approx 0.388 ~seconds}$

Learn more

• Velocity of A Runner : brainly.com/question/3813437
• Kinetic Energy : brainly.com/question/692781
• Acceleration : brainly.com/question/2283922

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Speed , Distance , Acceleration , Time , Velocity , Jet , Plane , TakeOff , Runway

Answer 2

speed of balloon which is rising upwards with constant speed is

$v_1 = 1.6 m/s$

speed of the camera which is thrown up from the ground

$v_2 = 10.2 m/s$

the distance between camera and balloon when it is thrown from the ground

$d = 2.60 m$

now let say after time "t" the camera is catch by her friend in the balloon

So the position of balloon and camera must be same

$y_{balloon} = y_{camera}$

$2.6 + 1.60 t = 10.2* t + \frac{1}{2}(-g)t^2$

$2.6 = 8.6 *t - 4.9 t^2$

Solving above quadratic equation we got

$t = 0.39 seconds$

So after t = 0.39 s her friend will get her camera on the balloon.