Question

john is riding the giant drop at canada if john free falls for2.6 seconds,what will be his final velocity and how far will he fall

Answer 1

Answer:

By using the first equation of motion:

V =u +at    

Here initial velocity is zero "0"

Put the values.

V = 0 + gt   (since g = 9.8 m/s2)

V = gt = 9.8 x 2.6  

         = 25.48 m/s  

Now find the distance by using the 2nd equation of motion.

S = ut+(1/2)at2       (Since u = o)

S = (1/2)gt2      

S= (1/2) x 9.8 x (2.6)2

S= 9.8 x 6.76 /2

S = 66.248 / 2

S= 33. 124 m

Answer 2

Answer:

By using the first equation of motion:

V =u +at    

Here initial velocity is zero "0"

Put the values.

V = 0 + gt   (since g = 9.8 m/s2)

V = gt = 9.8 x 2.6  

         = 25.48 m/s  

Now find the distance by using the 2nd equation of motion.

S = ut+(1/2)at2       (Since u = o)

S = (1/2)gt2      

S= (1/2) x 9.8 x (2.6)2

S= 9.8 x 6.76 /2

S = 66.248 / 2

S= 33. 124 m

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