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for a projectile launched upwards from level ground if the speed of a projectile at the maximum height is half the speed of projection u its horizontal range is
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clue v0x=v0/2
at the maximum height, the vertical component of velocity becomes zero. so only the horizontal component (u costheta) is present. from the given info. , \[u \cos \theta = u/2\] \[\cos \theta = 1/2\] \[\theta = 60\]
solve it on your own from here. and i'd rather suggest that you go through the theory part of the chapter before starting with the numerical problems
Write your solution here if it is not correct, we will help you.
i got it thanxxx guys
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