Question

please solve this problem.

Answer 1

hope it helps u........

Answer 2

Since the question arises from projectile motion (2 - directional ; xy plane) and not kinematics(1-directional; either x or y plane).. ..

Thus, the formula here used must be of projectile motions and not kinematics.

Note:-
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A projectile should have an angle else the range be 0. And thus then it becomes 1 - directional or linear.
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Hope you like my process
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=> gravity (g) = 10 m/s²

=> Let the velocity of projectile be v m/s

=> Velocity of cart (u) = 30 m/s

=> distance covered (x) = 80 m

=> time taken (T) $= \it\frac{x}{u} = \frac{80}{30} = \bf \frac{8}{3} \: \: sec$

Thus the total time taken by projectile will also be equal to the time taken by cart to reach the cart again.

Thus,

$= > t = \frac{2v \sin( \theta ) }{g} \\ \\ = > \frac{8}{3} = \frac{2 \times v \sin( \theta ) }{10} \\ \\ = > \bf v \sin( \theta ) = \underline{ \frac{40}{3} \: m {s}^{ - 1} }$

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Note:-
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The velocity of horizontal component of projectile should be same to the velocity of cart to reach the same point together.

Hence, relative horizontal component of velocity becomes 0.

Cart doesn't have any vertical component. So, velocity of projectile vertically will be the relative velocity of projectile seen by cart.
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Now..the answer,
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Horizontally, (not required)
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The particle seems moving with same velocity as the cart moves.

Vertically,
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The relative velocity of projectile to cart will be the vertical component of cart
$i.e \: \: \bf \frac{40}{3} \: \: m {s}^{ - 1}$
Thus,

$\underline{ \it \: the \: cart \: \: sees \: \: the \: \: projectile \: \: moving} \\ \ \underline{ \it \:velocity \: \: with \: \: a \: \: velocity \: \: of \: \: \bf \: \frac{40}{3} m {s}^{ - 1} }$
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Hence option c (✔️) is the required answer.
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Hope this is ur required answer

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