Question

A basketball is to throw in a basket which is situated at a height of 3 m at a horizontal distance 4 m from the point of throw. with minimum speed can the ball be projected to achieve the target

Answer 1

The minimum speed with which the ball is projected to achieve the target is 4 m/s

Explanation:

Let the speed of the projection be $u$ and angle of projection $\theta$

If the time taken is t

Then

$x=u\cos\theta\times t$

$y=u\sin\theta t-\frac{1}{2}gt^2$

Here,

$x=4$ m

$y=3$ m

Therefore,

$3=u\sin\theta\times\frac{4}{u\cos\theta}-\frac{1}{2}g(\frac{4}{u\cos\theta})^2$

$\implies 3=4\tan\theta-\frac{16}{u^2\cos^2\theta}$

$\implies \frac{16}{u^cos^2\theta}=4\tan\theta-3$

$\implies u^2\cos^2\theta=\frac{16}{4\tan\theta-3}$

$\implies u^2=\frac{16}{\cos^2\theta(4\tan\theta-3)}$

$\implies u^2=\frac{16}{4\sin\theta\cos\theta-3\cos^2\theta}$

Velocity will be minimum when $4\sin\theta\cos\theta-3\cos^2\theta$ is maximum

$4\sin\theta\cos\theta-3\cos^2\theta$

$=2\sin2\theta-3\cos^2\theta$

$=2\sin2\theta-3(\frac{\cos2\theta+1}{2})$

$=\frac{4\sin2\theta-3\cos2\theta-3}{2}$

We know that the maximum value of $a\sin\theta+b\cos\theta$ is $\sqrt{a^2+b^2}$

Therefore, the maximum value of $4\sin2\theta-3\cos2\theta$ will be $\sqrt{4^2+(-3)^2}=5$

Thus,

The maximum value of $\frac{4\sin2\theta-3\cos2\theta-3}{2}$ is $\frac{5-3}{2}=1$

Therefore, minumum value of $u^2$

$u^2_{min}=\frac{16}{1}$

$\implies u_{min}=\sqrt{16}=4$ m/s

Hope this answer is helpful.

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