A ball is projected from the ground level with the velocity of 72 km-hr at an angle of projection of 60 de -8203- -8203-rgee nbsp-with the horizontal find its position and velocity after 1 s-plezzzz help

U #160;= #160;72 #160;km/h #160;= #160;20 #160;m/sux #160;= #160;ucos60 #160;= #160;10 #160;m/suy #160;= #160;usin60 #160;= #160;1 ...

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Byju's Answer

Standard X

Physics

Escape Velocity

A ball is pro...

Question

A ball is projected from the ground level with the velocity of 72 km/hr at an angle of projection of 60 dergee with the horizontal find its position and velocity after 1 s.
plezzzz help

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Solution

$u = 72 k m / h = 20 m / s u_{x} = u \cos 60 = 10 m / s u_{y} = u \sin 60 = 10 \sqrt{3} m / s \sin c e t h e r e i s n o a c c e l e r a t i o n i n x d i r c t i o n v_{x} = u_{x} a n d a_{y} = - g v_{y} = u_{y} - g t = 10 \sqrt{3} - 10 = 10 (\sqrt{3} - 1) m / s v = \sqrt{{v_{x}}^{2} + {v_{y}}^{2}} = \sqrt{100 + 100 (3 + 1 - 2 (\sqrt{3}))} = 10 \sqrt{5 - 2 \sqrt{3}} = 12.39 m / s d i s \tan c e t a r v e l l e d s_{x} = u_{x} t = 10 \times 1 = 10 m s_{y} = u_{y} t - \frac{1}{2} g t^{2} = 10 \sqrt{3} - 0.5 \times 10 = 12.32 m s = \sqrt{{s_{x}}^{2} + {s_{y}}^{2}} = 15.86 m a w a y f r o m t h e i n i t i a l p o s i t i o n .$

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A ball is projected from the ground level with the velocity of 72 km/hr at an angle of projection of 60 de​​rgee with the horizontal find its position and velocity after 1 s. plezzzz help

A ball is projected from the ground level with the velocity of 72 km/hr at an angle of projection of 60 dergee with the horizontal find its position and velocity after 1 s.
plezzzz help