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in a shot out event a 2kg of shot is launched from a height of 1.50m, with an initial velocity of 8m/s at an angle of 60 degrees to horizontal
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what is the maximum height from the ground that is reached by the shot ?
by using the vertical component of velocity you can calculate h \[v^2 =u^2 -2gh\] at maximum height (h) v=0 ang u= 8 sin 60 calculate h and add 1.5 you will get your final answer. you can solve it by energy consevation.
i got an answer of 2.45m ?
add 1.5 in this and you will get your final answer.
so how do we find speed at the max height ?
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speed always zero at max height.
y?
it is condition when vertical velocity of object is 0 that's the max height it achieve.
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