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A boy throws a ball with initila velocity u at an angle of porjection theta from a tower of height H. Neglecting air resistance, find (a) how high above the building the ball rise, and (b) its speed just before it hits the ground.

Answer» <html><body><p></p>Solution :(a) Only gravitational force acts on the ball, which is conservative, therefore we can apply <a href="https://interviewquestions.tuteehub.com/tag/conservation-929810" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVATION">CONSERVATION</a> of energy i.e., U = 0. At the topmost point in the path, the ball is moiving horizonrally with <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> `ucos theta`. <br/> Initial total mechanical energy, <br/> `E_(i) = 0 + (1)/(2) m u^(2)` <br/> Total mechanical energy at thetopmost point : <br/> `E_(r) = (1)/(2) m u^(2) cos^(2)theta + mgh` <br/> From conservation of energy, we have `E_(i) = E_(r)` <br/> (a) `(1)/(2) m u^(2) = (1)/(2) <a href="https://interviewquestions.tuteehub.com/tag/mu-566056" style="font-weight:bold;" target="_blank" title="Click to know more about MU">MU</a>^(2) cos^(2) theta + mgh` <br/> or `h = (u^(2) - u^(2) cos^(2) thea)/(2g)` <br/> (b) If v is the <a href="https://interviewquestions.tuteehub.com/tag/speed-1221896" style="font-weight:bold;" target="_blank" title="Click to know more about SPEED">SPEED</a> of the ball at the ground, <br/> `E_(f) = (1)/(2) mv^(2) - mgh` <br/> From conservation of energy, wehave <br/> `E_(i) = E_(f) <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a>(1)/(2)m u^(2) = (1)/(2)mv^(2)-mgH.v=sqrt(u^(2)2gH)`</body></html>


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