Question

An object is launched at a 60 degree angle with an initial velocity of 100

m/s. What is the objects height? How long is the object in the air? What is
the objects's range (distance)? *

PLS HELP QUICK LOL

Answer 1

Step-by-step explanation:

The first question asks for the object's height. That is a y-dimension thing so we will use the one-dimensional formula:

`v^2=v_0^2+2a`Δx and we are looking for delta x, which is the displacement of the object (here, the height since displacement in the y-dimension is height). Before we do that, though, we need to know the initial upwards velocity, which is found in:

`v_(0y)=100sin(60)\\v_(0y)=90(m)/(s)`

And we also have to know from previous physics experience that the final velocity of an object at its max height is 0. Filling in:

`0=(90)^2+2(-9.8)`Δx

`0=8000-20`Δx and

-8000 = -20Δx so

Δx = 400 m

Now for the time it was in the air...we will use the simple equation

`v=v_0+at\\0=90-9.8t\\-90=-9.8t\\t=9`

So the object is in the air for 9 seconds. Now we go to the other dimension, the x-dimension, to find out how far it goes horizontally. We will use the equation:

Δx =
`v_0t+(1)/(2)at^2` but first we have to find the horizontal velocity, which is NOT the same as the upwards velocity.

`v_(0x)=100cos(60)\\v_(0x)=50(m)/(s)`

And another thing we need to know prior to solving problems like this is that the acceleration in the x-dimension is ALWAYS 0. Filling in the equation:

Δx =
`50(9)+(1)/(2)(0)(9)^2`

Δx = 500 m