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The maximum height attained by a projectileis increased by 5%. Keeping the angle of projection constant, what is the percentage increase in horizontal range?
  • a)
    5 %
  • b)
    10 %
  • c)
    15 %
  • d)
    20 %
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The maximum height attained by a projectileis increased by 5%. Keeping...
Given: The maximum height attained by a projectile is increased by 5%.

To find: The percentage increase in horizontal range, keeping the angle of projection constant.

Solution:

Let's assume that the initial maximum height attained by a projectile is H and the initial horizontal range is R. After increasing the maximum height by 5%, the new maximum height attained by the projectile is 1.05H.

Now, we know that the horizontal range of a projectile is given by the formula:

R = (u^2 * sin2θ)/g

where u is the initial velocity of the projectile, θ is the angle of projection, and g is the acceleration due to gravity.

Since the angle of projection is constant, we can write:

R ∝ u^2

This means that the horizontal range is directly proportional to the square of the initial velocity of the projectile.

Now, let's consider two cases:

Case 1: Initial velocity and maximum height remain constant

In this case, the initial velocity and the angle of projection remain constant. Therefore, the horizontal range is also constant. Hence, the percentage increase in the horizontal range is 0%.

Case 2: Initial velocity remains constant, but the maximum height increases by 5%

In this case, the initial velocity remains constant, but the maximum height increases by 5%. Therefore, the time of flight of the projectile also increases, as the projectile takes more time to reach the new maximum height. This, in turn, increases the horizontal range.

Let the new horizontal range be R'. Then, we can write:

R' = (u^2 * sin2θ)/g'

where g' is the new acceleration due to gravity. Since the maximum height attained by the projectile is directly proportional to the square of the initial velocity, we can write:

1.05H ∝ u^2

or u^2 = (1.05H)/k

where k is a constant of proportionality.

Substituting this value of u^2 in the formula for horizontal range, we get:

R' = [(1.05H * sin2θ)/k]/g'

or R' = (1.05sin2θ * R)/g'

Therefore, the percentage increase in the horizontal range is:

[(R' - R)/R] * 100

= [(1.05sin2θ * R)/g' - R]/R * 100

= (0.05sin2θ * R)/R * 100

= 5%

Hence, the correct answer is option A) 5%.
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The maximum height attained by a projectileis increased by 5%. Keeping the angle of projection constant, what is the percentage increase in horizontal range?a)5 %b)10 %c)15 %d)20 %Correct answer is option 'A'. Can you explain this answer?
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