A javelin thrown into air at an angle with the horizontal has a range of $200 m$ . If the time of flight is $5 s$ , then the horizontal component of velocity of the projectile at the highest point of the trajectory is

40

m s^{- 1}

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0

m s^{- 1}

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9.8

m s^{- 1}

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\infty

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Solution

The correct option is B $40$ $m s^{- 1}$
Range: $S = u c o s θ \times T$
$∴ u cos θ = \frac{200}{5} = 40 m / s$
The horizontal component of velocity $u cos θ$ remains same throughout journey.

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A javelin thrown into air at an angle with the horizontal has a range of 200m. If the time of flight is 5s, then the horizontal component of velocity of the projectile at the highest point of the trajectory is

The correct option is B 40 ms−1Range: S=ucosθ×T∴ucosθ=2005=40m/sThe horizontal component of velocity ucosθ remains same throughout journey.

A javelin thrown into air at an angle with the horizontal has a range of $200 m$ . If the time of flight is $5 s$ , then the horizontal component of velocity of the projectile at the highest point of the trajectory is

The correct option is B $40$ $m s^{- 1}$
Range: $S = u c o s θ \times T$
$∴ u cos θ = \frac{200}{5} = 40 m / s$
The horizontal component of velocity $u cos θ$ remains same throughout journey.