Question

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Answer 1

★Given:-

• Velocity of cart = 30 m/s.

• A projectile is to be fired from the moving cart in such a way that it will return to the cart after the cart has moved 80 m.

★To find:-

• At what speed (Relative to the cart) must the projectile be fired?

★Solution:-

According to question,

Car is moving in the x axis as it is moving horizontally.

The projectile the moves vertically upward and comes back.

We know:

$\leadsto \underline{\boxed{\sf Time = \frac{Distance}{speed} }}$

The time taken by cart to cover 80 m

:-

$:\implies \sf t=\dfrac{d}{v} \\\\:\implies \sf t=\dfrac{80}{30} \\\\:\implies \bold{t=\dfrac{8}{3} s.}$

If a projectile is moving upward,

a = - g

v = 0

And,

$\sf t=\dfrac{(8/3)}{2}$

We know:

$\mapsto \underline{\boxed{\sf v = u+at}}$

Where,

v = final velocity

u = initial velocity

a = acceleration

t = time

Substituting the values in this formula,

$:\implies \sf 0 = u+(-10)(\frac{4}{3} )\\\\:\implies \sf 0=u-10\times \frac{4}{3} \\\\:\implies \sf 0 = u-\frac{40}{3} \\\\:\implies \boxed{\boxed{\sf u = \frac{40}{3}m/s }}$

Hence,

Speed (Relative to the cart) at which the projectile must be fired is 40/3m/s.

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Answer 2

Answer:

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Explanation: