Question

Example 8.
Find the value of o(teta) in the diagram given below so that the projectile can hit the target.

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Answer 1

Answer:

i mean your question is incomplete man...

Answer 2

Answer:

The value of θ so that the projectile can hit the target is 45° or tan⁻¹(3).

Explanation:

To find: Find the value of θ in the diagram given below so that the projectile can hit the target.

Solution:

From equation of trajectory we have,

$y = x\tan\theta - \dfrac{gx^2(1+\tan^2\theta)}{2u^2}\\\\10 = 20\tan\theta-\dfrac{5\times20^2(1+\tan^2\theta)}{20^2}\\\\\tan^2\theta-4\tan\theta+3=0\\\\(\tan\theta-3)(\tan\theta-1)=0\\\\\tan\theta=1,3\\\\\boxed{\theta = 45^o \ or \ \theta = \tan^{-1}(3)}$

Therefore, The value of θ so that the projectile can hit the target is 45° or tan⁻¹(3).