Answer:
Explanation:
The first question asks for the object's height. That is a y-dimension thing so we will use the one-dimensional formula:
[tex]v^2=v_0^2+2a[/tex]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:
[tex]v_{0y}=100sin(60)\\v_{0y}=90\frac{m}{s}[/tex]
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:
[tex]0=(90)^2+2(-9.8)[/tex]Δx
[tex]0=8000-20[/tex]Δx and
-8000 = -20Δx so
Δx = 400 m
Now for the time it was in the air...we will use the simple equation
[tex]v=v_0+at\\0=90-9.8t\\-90=-9.8t\\t=9[/tex]
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 = [tex]v_0t+\frac{1}{2}at^2[/tex] but first we have to find the horizontal velocity, which is NOT the same as the upwards velocity.
[tex]v_{0x}=100cos(60)\\v_{0x}=50\frac{m}{s}[/tex]
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 = [tex]50(9)+\frac{1}{2}(0)(9)^2[/tex]
Δx = 500 m
