Question

Question #50ba8

Answer 1

The stone is 17.3 m away in the horizontal direction, and 13.4 m above the ground.

Explanation:

Here's what you must do:

Since the velocity given in the problem is an what we call "standard form" - its magnitude and direction are stated - you must convert it into "rectangular form" (also known as component form) so that you can separate the horizontal aspect of the motion from the vertical. Once this is done, the horizontal and vertical components of the displacement will be easier to obtain.

The equations you need are

`v_(0x) = v_0 cos theta`

where I use `v_o` to represent the initial velocity and `theta` is the launching angle.

Likewise

`v_(0y)=v_0 sin theta`

In the above example

`v_(0x)= 26 cos 42° = 19.2` m/s, and `v_(0y)=26 sin 42° = 17.4` m/s

Now, use equations of motion. For the horizontal displacement:

`x=v_(0x) t = 19.2 xx 0.90 = 17.3 m`

and for the vertical displacement:

`y=v_(0y)-1/2 g*t^2 = 17.4 - 4.9(0.90)^2=13.4 m`