Question

A gun crew observes a remotely controlled balloon

launching an instrumented spy package in enemy

territory. When first noticed the balloon is at an

altitude of 800 m and moving vertically upward at a

constant velocity of 5 m/s and 1600 m away. Shells

fired from the gun have an initial velocity of

400 m/s at a fixed angle (sin = 3/5 and

cos = 4/5). The gun crew waits and fires so as

destroy the balloon. Assume g = 10 m/s2. Neglect air

resistance

1. What is the flight time of the shell before it strikes

the balloon (in sec) ?

2. What is the altitude of the collision (in meter)?

3. How long did the gun crew wait before they fired

(in sec) ?​

Answer 1

Your Question:-

⇒ A gun crew observes a remotely controlled balloon  

⇒ A gun crew observes a remotely controlled balloon  launching an instrumented spy package in enemy  territory. When first noticed the balloon is at an  altitude of 800 m and moving vertically upward at a  constant velocity of 5 m/s and 1600 m away. Shells  

away. Shells  fired from the gun have an initial velocity of  

400 m/s at a fixed angle (sin = 3/5 and  

(sin = 3/5 and  cos = 4/5). The gun crew waits and fires so as  destroy the balloon. Assume g = 10 m/s^2. Neglect air  

Neglect air  resistance

1. What is the flight time of the shell before it strikes

the balloon (in sec) ?

2. What is the altitude of the collision (in meter)?

3. How long did the gun crew wait before they fired

(in sec) ?

                                 

Your Answer:-

1) = 5 sec

2) = 1075m

3) = 50 sec

                                 

Given :-

sinθ = 3/5

cosθ = 4/5

u = 400m/s

s = 800m

v = 5m/s

g = 10m/s^2

(please note :- this Question is totally based of figure bue i cant draw the digram here so it can be atachhed of the bottem or top of your screen so please see that digram..)

To find :-

1. What is the flight time of the shell before it strikes

the balloon (in sec) ?

2. What is the altitude of the collision (in meter)?

3. How long did the gun crew wait before they fired

(in sec)

Solution:-

1) the motion in the x-direction is a constant velocity motion.

We find the flight time = $\frac{1600}{v_{x} }$

=  $\frac{1600}{400cos\theta }$

=  $\frac{1600}{\frac{1600}{5} }$

=  $5 sec$

hence the time of flight  = 5 sec

                                   

2)  from the flight , the initial velocity in y-direction and the acceleration in the y-direction , we can calculate the altitude of the shell:

= $h = v_{y}t - \frac{1}{2} gt^{2}$

= $\frac{1200}{5} *5 - \frac{1}{2} *10 *25$

= $1200 - 125 = 1075m$

hence the altitude is 1075m.

                                   

3) After the waiting time plus the flight time , the balloon should reach the same altitude as the shell .

Let $t_{w}$ be the waiting time .

We have:-

= $h = ( t_{w} + 5 ) * 5 + 800 = 1075$

= $t_{w} + 5 = \frac{275}{5}$

= $= 55 sec$

So, $t_{w} = 50 sec$

hence the waiting time = 50 sec

Thank you

please dont report it it take my so much time

and it is 100% correct..

thank you...................