Given,
Initial launcher speed = v
If initial launcher speed =3v
a) Let, initially the maximum height was h, and finally it was H
Applying newton's law of motion,
"v^2= u^2-2gh"
Now, at the maximum height v=0
"\\Rightarrow 0=v^2-2gh"
"h=\\frac{v^2}{2gh}"
and finally "H=\\frac{9v^2}{2g}"
b)
Let the time of hang is T, and time taken to reach at maximum height = t
Now, "v=u-gt"
at the top most point, final velocity of the object = 0
"\\Rightarrow 0=v-gt"
"\\Rightarrow t= \\frac{v}{g}"
Hence, initially the time of flight of the projectile "(T)=2t = \\frac{2v}{g}"
and for the final condition,
"T' = \\frac{6v}{g}"
c)
The impact speed will be v if it was projected with v and it will be 3v, if it is projected with 3v.
but if "3v\\ge V_{esc}" ,
then it will not come back in the gravitational field.
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