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BartSMP [9]
3 years ago
9

An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou

nd. A friend picks it up and throws it upward at 20 m/s at the instant the passenger is 5 m above the ground. (10 pts) a) Calculate the time for camera to reach passenger. b) Calculate the position and velocity of the camera when passenger catches it.
Physics
1 answer:
frosja888 [35]3 years ago
8 0

Answer:t=0.3253 s

Explanation:

Given

speed of balloon is u=3\ m/s

speed of camera u_1=20\ m/s

Initial separation between camera and balloon is d_o=5\ m

Suppose after t sec of  throw camera reach balloon then,

distance travel by balloon is

s=ut

s=3\times t

and distance travel by camera to reach balloon is

s_1=ut+\frac{1}{2}at^2

s_1=20\times t-\frac{1}{2}gt^2

Now

\Rightarrow s_1=5+s

\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t

\Rightarrow 5t^2-17t+5=0

\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}

\Rightarrow t=\dfrac{17\pm 13.747}{10}

\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is  t=0.3253\ s

velocity is given by

v=u+at

v=20-10\times 0.3253

v=16.747\ m/s

and position of camera is same as of balloon so

Position is =5+3\times 0.3253

=5.975\approx 6\ m

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