When a scientist finds answers to questions brought up by scientific inquiry, he needs to:
1 answer:
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For this we will use kinematic equations.
We need the time of flight to compute the range.
With the given information, the most useful formula is:
Δy=v0yt−g2t2
Because our equation is taking only vertical motion into account,
v0y=30sin(45)
Because we are looking for the range of the object,
Δy=0
This leaves us with only one unknown: t
0=30sin(45)t−g2t2
g2t2=30sin(45)t
g2t=30sin(45)
t=60gsin(45)
For horizontal distance, the formula we need is:
Δx=v0xt
Because we are taking into account the horizontal range,
v0x=30cos(45)
In substituting our derived values into the equation we can see that:
Δx=30cos(45)∗60gsin(45)
Range=30cos(45)∗60gsin(45)
