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Grade: 11
        
A balloon starts rising from the ground with an acceleration of 1.25 metre per second square a stone is released from the balloon after 10 seconds determine maximum height of the stone from the ground and time taken by the stone to reach the ground
one month ago

Answers : (2)

Arun
21570 Points
							

After 10s the upward velocity of the balloon is =( according to u+at) (1.25)(10)=12.5 m/s.

Therefore, the stone will have initial upward velocity ,

u= 12.5 m/s………..(1)

We have taken upward direction as positive.

The acceleration of the stone, after it is realesed is

g=- 9.8 m/s^2………..(2)

The height of the balloon at 10s is ,

h=(1/2)at^2=(1/2)(1.25)(100)=62.5 m.

Now, using h=ut+(1/2)gt^2, we have

-40=12.5t- (1/2)(9.8)t^2. Therefore,

No you can solve this

one month ago
Khimraj
2914 Points
							
 
Steps:
1) Initially, 
Initial velocity, u = 0m/s 
acceleration, a =1.25 m/s^2 
Velocity of balloon after 10 s, v = u +at 
=> v = 0 + 1.25 * 10 = 12.5m/s 
2) Displacement of balloon in 8 s, 
s = ut + \frac{1}{2} a {t}^{2} \\ = > 0 + \frac{1}{2} \times 1.25 \times {10}^{2} = 62.5m
3) After t = 8s, a stone is dropped. 
Initial velocity of stone at time of fall is same as that of balloon at that time :
That is velocity of stone,when it is dropped is 10m/s upward. 
Initial velocity, u = 12.5m/s
Displacement, s = -62.5m ( - means downward) 
Balloon is left. 
Acceleration, a = -10 m/s^2 

So, time taken by stone to reach the
s = ut + \frac{1}{2} a {t}^{2} \\ - 62.5 = 10t + \frac{1}{2} \times( - 12.5) \times {t}^{2} \\
 
now solve for t
 
one month ago
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