A cart is moving horizontally along a straight line with constant speed $30 m/s$ . A particle is to be fired vertically upwards from the moving cart in such a way, that it returns to the point of projection on the cart after cart has moved $80 m$ . At what speed (relative to cart) must the particle be projected?

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Solution

Answer (3)

$R = \frac{2 u_{x} u_{y}}{g} \Rightarrow 80 = \frac{2 (30) (u_{y})}{10}$
$u_{y} = \frac{80}{6} = \frac{40}{3} {ms}^{- 1}$

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Answer (3) R=2uxuyg⇒80=2(30)(uy)10 uy=806=403ms−1

Answer (3)

$R = \frac{2 u_{x} u_{y}}{g} \Rightarrow 80 = \frac{2 (30) (u_{y})}{10}$
$u_{y} = \frac{80}{6} = \frac{40}{3} {ms}^{- 1}$