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A projectile A is projected from ground....

A projectile A is projected from ground. An observer B running on ground with uniform velocity of magnitude 'v' observes A to move along a straight line. The time of flight of A as measured by B is T. Then the the range R of projectile on ground is

A

`R=vT`

B

`RltvT`

C

`RgtvT`

D

information insufficient to draw inference

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The correct Answer is:
To solve the problem, we need to find the range \( R \) of the projectile \( A \) as observed from the ground, given that an observer \( B \) running with a uniform velocity \( v \) sees \( A \) moving in a straight line and measures the time of flight \( T \). ### Step-by-Step Solution: 1. **Understanding the Motion of the Projectile**: - The projectile \( A \) is launched from the ground with an initial velocity \( u \) at an angle \( \theta \) with respect to the horizontal. - The horizontal component of the projectile's velocity is given by \( u \cos \theta \). - The vertical component of the projectile's velocity is given by \( u \sin \theta \). 2. **Time of Flight**: - The total time of flight \( T \) for the projectile is given by the formula: \[ T = \frac{2u \sin \theta}{g} \] where \( g \) is the acceleration due to gravity. 3. **Observer's Perspective**: - Observer \( B \) is running with a uniform velocity \( v \) and sees the projectile moving in a straight line. For this to happen, the horizontal component of the projectile's velocity must equal the velocity of observer \( B \): \[ u \cos \theta = v \] 4. **Calculating the Range**: - The range \( R \) of the projectile on the ground can be calculated using the horizontal component of the velocity and the time of flight: \[ R = \text{(horizontal velocity)} \times \text{(time of flight)} = (u \cos \theta) \times T \] - Substituting \( u \cos \theta = v \) into the equation gives: \[ R = v \times T \] 5. **Final Expression**: - Thus, the range \( R \) of the projectile as measured from the ground is: \[ R = vT \] ### Conclusion: The range \( R \) of the projectile on the ground is given by: \[ R = vT \]

To solve the problem, we need to find the range \( R \) of the projectile \( A \) as observed from the ground, given that an observer \( B \) running with a uniform velocity \( v \) sees \( A \) moving in a straight line and measures the time of flight \( T \). ### Step-by-Step Solution: 1. **Understanding the Motion of the Projectile**: - The projectile \( A \) is launched from the ground with an initial velocity \( u \) at an angle \( \theta \) with respect to the horizontal. - The horizontal component of the projectile's velocity is given by \( u \cos \theta \). - The vertical component of the projectile's velocity is given by \( u \sin \theta \). ...
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