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| Maximum Height |
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| Maximum height is denoted by the letter hmax or H. It is also known as the vertical range. It is the maximum height to which a projectile rises above the horizontal plane of projection. In order to calculate the maximum height H, we make use of the fact that the velocity vy(t) of the projectile at the maximum height is zero. If t is the time taken by the projectile to reach maximum height, then from the equation
vy = v sin q - gt |
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| We have |
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| At time t1, y (t) = H, the maximum height |
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| Substituting for t1, |
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| It is the total time taken by the projectile to return to the same level from where it was thrown. |
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| Since the time of ascent is equal to the time of descent, the time of flight is equal to twice the time taken by the projectile to reach the maximum height. |
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| Therefore, time of flight, T = 2t where t is the time of ascent or time taken by the projectile to reach the maximum height. Once again, |
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 In the equation, |
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| We have, |
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| Hence, the time of flight, T = 2t |
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| It is the total horizontal distance from the point of projection to the point where the projectile returns back to the plane of projection. It is denoted by R. |
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| To calculate the horizontal range, we need to consider the horizontal motion of the uniform velocity vcosq, in the horizontal direction. Hence, the horizontal range is given by, |
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Note: R = R maximum, when  |
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| Horizontal range is give by |
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| This shows that there are two angles of projection for the same
horizontal range i.e., J and (90o
- J) with the horizontal. The projectile will cover the same horizontal range whether it is thrown at an angle (900-q) with the horizontal, or an angle q with the vertical. |
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