Question

If a stone is to hit at a point which is at a distance d away and at a height h above the point from where the
stone starts as shown in the figure then what is the value of initial speed u if stone is launched at an theta ?

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Answer 1

Given :

▪ Initial velocity of projection = u

▪ Angle of projection = $\sf{\theta}$

▪ Horizontal distance of point from origin = d

▪ Vertical distance of point from ground = h

To Find :

▪ Initial velocity of projection.

Formula :

↗ Equation of trajectory :

$\bigstar\:\underline{\boxed{\bf{\red{y=x\tan\theta-\dfrac{gx^2}{2u^2\cos^2\theta}}}}}$

• y denotes verticle distance
• x denotes horizontal distance
• g denotes acceleration due to gravity
• $\sf{\theta}$ denotes angle of projection

Calculation :

$\dashrightarrow\bf\:y=x\tan\theta -\dfrac{gx^2}{2u^2\cos^2\theta}\\ \\ \dashrightarrow\sf\:h=d\tan\theta-\dfrac{gd^2}{2u^2\cos^2\theta}\\ \\ \dashrightarrow\sf\:\dfrac{gd^2}{2u^2\cos^2\theta}=d\tan\theta-h\\ \\ \dashrightarrow\sf\:\dfrac{1}{u^2}=\dfrac{2cos^2\theta(d\tan\theta-h)}{gd^2}\\ \\ \dashrightarrow\underline{\boxed{\bf{\blue{u=\dfrac{d}{\cos\theta}\sqrt{\dfrac{g}{2(d\tan\theta-h)}}}}}}\:\orange{\bigstar}$

✴ Extra Dose :

• A body is said to be projectile is it is projected into space with some initial velocity and then it continues to move in a vertical plane such that its horizontal acceleration is zero and vertical downwards acceleration is equal to g.

Answer 2

$\huge{\pink{\underline{\underline{Question}}}}$

If a stone is to hit at a point which is at a distance D away and at a height h above the point from where the stone starts as shown in the figure then what is the value of Initial speed u if the stone is launched as an theta.

$\huge{\purple{\underline{\underline{Answer}}}}$

Option.B

Explanation:-

Given,

➡️Initial velocity = u

➡️Angle = theta

➡️Horizontal distance of a point = d

➡️Height = h(Vertical distance)

We have to find the initial velocity of a position.

Formula :-

$y = x \: tan(theta) - \frac{ {gx}^{2} }{ {2u}^{2} {cos}^{2} theta}$

$h = d \: tan(theta) - \frac{ {gd}^{2} }{2 {u}^{2} {cos}^{2}theta }$

$\frac{ {gd}^{2} }{ {2u}^{2} \: {cos}^{2} theta } = d \: tan \: (theta) - h$

$\frac{1}{ {u}^{2} } = \frac{2 {cos}^{2} theta \: - h}{ {gd}^{2} }$

$u = \frac{d}{cos(theta)} \sqrt{ \frac{g}{2(d \: tan \: (theta) - h} }$

so, the answer of your question is Option.B