Question

A cannonball is shot form a cliff 500 feet high with an initial velocity of 20 ft/s. If the angle of the ball with the horizontal is 45 degrees, when does the ball hit the ground? How far does the ball go horizontally?

Answer 1

$\blacksquare\,\,\,\underline{Projectile\,\, Motion}:$

2ocos 450 720 ftols vsinuso 450 45 ) 2ocos (vcosys t. 20 cos 450 500 d

DIAGRAM:

$Let\,\,the\,\,ball\,\,take\,\,t\,\, second\,\,time\,\,to\,\,hit\,\, the\,\, ground\,.\\And\,\,the\,\,ball\,\,go\,\,the\,\,distance\,\, horizontally ,\,\,d\,\,ft.\,\,.\\\\Given,\,\,the\,\, initial\,\,height\,\,of\,\,the\,\,ball \,\,is,\,\,500\,\,ft.\,\,and\\initial\,\, velocity\,\,20ft./sec.\,\,and\,\, angle\,\,with\,\, horizontal\\is,\,\,45\degree\,\,.\\\\We\,\, know,\,\,at\,\,the\,\,point,\,\,the\,\,ball\,\,just\,\,touched\,\,the\\ ground,\,\,the\,\, vertical\,\, displacement\,\,will\,\,be\,\,0\,\,.\\\\ So,\,\,from\,\,the\,\,formula,\,\,s=ut+\frac{1}{2}at^{2}+h\,,we\,\,have,\\\\ \Rightarrow\,\,0=20sin\,45\degree\times t-\frac{1}{2}\times g\times t^{2} +500\,\,[Here,\\\\u=vertical\,\, velocity\,\,=20 sin45\degree,\,\,and\,\,h=500ft.\,]\\\\\Rightarrow\,\, t\,=\frac{20sin\,45\degree+\sqrt{(20 sin\,45\degree)^{2}+2\cdot g\cdot\,500}}{g} \\\\\Rightarrow\,\,t\,=\frac{20\,sin45\degree+\sqrt{32374.1}}{32.1741} \,\,\,[\,Here,\,\,g=32.1741\,\,ft./sec.^{2}]$

$\Rightarrow\,\,t\,=6.03946134\,\approx 6.04\,\,sec.\,.\\\\Next,\,\,we\,\,have\,\,to\,\,find\,\,the\,\, horizontal\,\, displacement \\d\,\,.\,\,For\,\,that,\,\,using\,\,the\,\,formula\\s=ut+\frac{1}{2} at^{2}\\\\Since,\,\,there\,\,is\,\,no\,\, acceleration\,\,to\,\,the\,\, horizontal\,\, \\so,\,\,a=0,\\\\s=ut\\\\\Rightarrow\,\,d\,=20cos45\degree \times 6. 04\,\,[\,The\,\, horizon tal\,\, velocity=20cos45\degree\\\\and\,\,time=t=6.04\,] \\\\\Rightarrow\,\,d\,=85.41088812\approx 85.41\,\,[\,in\,\,two\,\,decimal\,] \\\\\therefore\,After\,\,6.04\,sec.\,\,the\,\,ball\,\,will\,\, hit\,\,the\,\, ground\,.\,\,And\,\,the\\ horizontal\,\, displacement\,\,of\,\,the\,\,ball\,\,is,\,\,85.41\,ft.\,\,.$

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