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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
5 months ago

Answers : (2)

Arun
22784 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

5 months ago
Khimraj
3008 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
 
5 months ago
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