Question

A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?

Answer 1

Answer:53.6722

Explanation:If anyone is taking Acellus, trust me, this is the answer.

Answer 2

Answer:53.63
`ms^(-2)`

Step-by-step explanation:

The equations of motion used in this question is
`v=u+at`

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of
`9.8ms^(-2)`.

In X-Direction,

Given that initial velocity=
`u_(x)`=
`33.8ms^(-1)`

Using
`v=u+at`,

`v_(x)=33.8+(0)4.25=33.8ms^(-1)`

In Y-Direction,

Given that initial velocity=
`u_(x)`=
`0ms^(-1)`

Using
`v=u+at`,

`v_(y)=0+(9.8)4.25=41.65ms^(-1)`

`v=\sqrt{v_(x)^(2)+v_(y)^(2)}`

`v=√(1142.44+1734.72)=√(2877.163)=53.63ms^(-1)`