John is riding the Giant Drop at Canada. If John free falls for 2.6 seconds, what will be his final velocity and how far will he fall?     In this question how u = 0??? V=0 ryt? because he is riding so he has motion. and after falling , he will come in rest ryt? Plz explain 

Dear Student,
Given,
John is undergoing a free fall. So, John will fall under the action of uniform acceleration g = 9.8 ms-2.
Also, initial velocity,  u = 0.
After t = 2.6 s
Let the velocity attained after t = 2.6 m/s be v. 
Then,
Using the relation:
v=u+at=gtv=9.8×2.6=25.48ms
Let it fall by a distance d.
Using another equation of motion:
d=ut+12at2=12gt2d=12×9.8×2.62=33.12 m

Regards

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The expression for the equation of the motion during free fall condition is as follows;

v= u+at

Here, u is the initial speed, v is the final velocity, g is the acceleration due to gravity and t is the time taken.

It is given in the problem that john is riding a giant drop at canada if john free falls for 2.6 s.

Put g= 9.8 meter per second square, t= 2.6 s and u = 0.

v= 0+(9.8)(2.6)

v= 25.48 m\s

Therefore, the final velocity is 25.48 m\s.

Calculate the distance.

Put u= 0, t= 2.6 s and g= 9.8 meter per second square.

Therefore, the distance is 33.12 m.

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