Question

When two projectiles are fired from complementary angles having times of flight T1 ,T2 and maximum height H1,H2. Find relation between R and H here H:height , R:range

Answer 1

Answer: R = 4√H1.H2

Explanation:Hope it helps !

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Answer 2

Given:

The angle of projectiles are complementary angles.

Time of flight for 1st projection = T1

Time of flight for 2nd projection = T2

Maximum height of 1st projection = H1

Maximum height for 2nd projection = H2

To find:

The relation between R and H.

Solution:

Let the angle of projection be α, β

Since the angles are complementary angles:

α + β = 90°

So we can take one angle as α and the other as 90°-α

Since height of a projectile:

H = u² sin ²α/ 2g

H1 = u² sin²α/ 2g

H2 = u² sin²(90°-α)/2g = u²cos²α/ 2g

We know that range of a projectile:

R= u² sin2α/ g

Now multiplying the heights we get:

H1 * H2 = (u²(sinα cosα))²/ 4g²

Replacing the value of R by R, we get:

H1* H2 = R²/ 16

R² = 16 H1 H2

Therefore the square of the range of the projectile is equal to 16 time the product of maximum heights of the projectile.