Question

The maximum range of a projectile is 22 m. When
it is thrown at an angle of 15® with the horizontal,
its range will be-​

Answer 1

Explanation:

the sum of 4th term and 8th term is 24and the s10thof 6th and 10th term is 44find AP

Answer 2

The range of the projectile will be 11 m.

Given,

Maximum range = 22 m.

Thrown at an angle = 15°.

To Find,

The range of the projectile.

Solution,

We have, Maximum range = 22 m

We know that, Range of projectile = v²sin2θ/g

∴ The range is maximum at 45°

⇒ v²sin2x45°/g = 22

⇒ v²/g = 22                                          --------------------------------------------(1)

Range of projectile at 15°,

⇒ v²sin2x15°/g = v²sin30°/g

                         = (v²/g)x sin30°

from (1),

⇒ 22 x sin30°

⇒ 22 x 1/2

⇒ 11 m.

Hence, the range of the projectile will be 11 m.

#SPJ3