Answer:
Δx=v∗t=(34.64 m/s)(4.19 s)=145.1m
Explanation:
First resolve the body’s original velocity into perpendicular components — horizontal and vertical.
The horizontal component is adjacent to the 30º angle, so we’ll set it up like this:
cos30º=vh40 m/s
Rearranging we get:
vh=(40 m/s)cos30º=34.64 m/s
(I’m not sure if you’re meant to round your answers to the appropriate number of significant figures or not, so I’ll just leave it like that and you can round as necessary.)
Next let’s calculate the vertical component of the body’s original velocity:
sin30º=vv40 m/s
vv=(40 m/s)sin30º=20 m/s
Now let’s figure out how long it will take to hit the ground. When it hits the ground, its vertical displacement will be 170 meters (treating down as positive). We know that its initial vertical velocity is 20 m/s, and the acceleration due to gravity is 9.81 m/s².
Δx=vi∗t+12at2
170 m=(20 m/s)t+12(9.81 m/s²)t2
We can rearrange this to get a quadratic equation:
0=(4.905 m/s²)t2+(20 m/s)t−170 m
Which you can solve with the quadratic formula. I’ll not type out the entire solution here, but you get two values for t:
t = -8.26 seconds and t = 4.19 seconds
We can discard the negative solution as extraneous. The body will hit the ground 4.19 seconds after being thrown.
Earlier we calculated that the body’s horizontal velocity is 34.64 m/s. Since the body’s horizontal velocity doesn’t change, we should be able to easily calculate its horizontal displacement during the 4.19 seconds of flight time.
