Question

john is riding a giant drop at canada if john free falls for 2.6 sec what will be his final velocity and how far he fall

Answer 1

Answer: The correct answers are 25.48 m\s and 33.12 m.

Explanation:

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.

$S=ut+\frac{1}{2}gt^{2}$

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

$S=(0)t+\frac{1}{2}(9.8)(2.6)^{2}$

$S=33.12 m$

Therefore, the distance is 33.12 m.

Answer 2

Answer:

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. Therefore, the final velocity is 25.48 m\s

Explanation:

hope it will help you