Question

A 2-kilogram block rests at the edge of a platform that is 10meters above level. The block is launched horizontally from theedge of the platform with an initial speed of 3 meters per second.Air resistance is negligible. The time it will take for the blockto reach the ground is most nearly

A. 0.3 s
B. 1.0 s
C. 1.4 s
D. 2.0 s
E. 3.0 s

Answer 1

The time it will take for the block to reach the ground is most nearly C. 1.4 s

Step-by-step explanation:

We have the following data in order to solve the exercise :

The mass ''m'' of the block

`m=2kg`

The high of the platform above a certain level which is
`10m`

We also know that the block is launched horizontally from the edge of the platform with an initial speed of
`3(m)/(s)`

This speed is exclusively horizontal. The vertical component of the velocity is
`0(m)/(s)`

Finally, the air resistance is negligible.

We can modeled this exercise as a free fall motion.

We only need this equation to solve the exercise :

`Y_(f)(t)=Y{i}+v{i}.t+(1)/(2).a.t^(2)` (I)

Where
`Y_(f)` is the final vertical position

Where
`Y_(i)` is the initial vertical position

Where
`v_(i)` is the initial speed

Where
`a` is the acceleration due to gravity which is
`9.81(m)/(s^(2))`

And where
`t` is the time variable

The equation (I) modelates the vertical position in function of the variable ''t'' time.

If we use the equation (I) with some simplifications such as
`Y_(i)=0m` and
`v_(i)=0(m)/(s)` because the initial vertical speed is zero, we find that

`10m=(9.81(m)/(s^(2)).t^(2))/(2)`

`t^(2)=(20m)/(9.81(m)/(s^(2)))`

`t^(2)=2.0387s^(2)`

`t=1.428s`

That is the time in which the block will travel
`10m` from the platform to the ground. The correct option is C. 1.4 s

Answer 2

C 1.4 s

Step-by-step explanation:

Even though we have a velocity on the horizontal axis it does not affect the time to reach the ground, so we can solve this as a free-fall motion:

`y=y_o-(1)/(2)*g*t^2\\t=\sqrt{(2*10m)/(9.8m/s^2)}\\t=1.4s`

so the correct answer is C, 1.4 seconds.