Derive the formula for Range of a projectile $(R)$. Derive the formula for maximum projectile.
Derive the formula for Range of a projectile $(R)$. Derive the formula for maximum projectile.
Range is defined as the horizontal distance travelled by a projectile from its initial position $y=0$ ) to the final position (where it passes $y=0$ during its fall) is called the Range of the projectile.
or it is the distance travelled from its initial position $(x=0, y=0)$ to final position $(x=\mathrm{R}, y=0)$ is called 'R'.
At any instant of time keeping $x=\mathrm{R}$ and $\mathrm{t}=\mathrm{t}_{\mathrm{F}}$ then $x=\left(v_{0} \cos \theta_{0}\right) t$
$\therefore \mathrm{R}=\left(v_{\mathrm{o}} \cos \theta_{\mathrm{o}}\right) t_{\mathrm{F}}$
$\therefore \mathrm{R}=\left(v_{\mathrm{o}} \cos \theta_{\mathrm{o}}\right)\left(\frac{2 v_{o} \sin \theta_{o}}{g}\right)\left[\because t_{\mathrm{F}}=\frac{2 v_{o} \sin \theta_{o}}{g}\right]$
$\therefore \mathrm{R}=\frac{2 v_{o}^{2} \sin \theta_{o} \cos \theta_{o}}{g}$
$\therefore \mathrm{R}=\frac{v_{o}^{2}\left(\sin \theta_{o} \cos \theta_{o}\right)}{g}$
$\therefore \mathrm{R}=\frac{v_{o}^{2} \sin 2 \theta_{o}}{g}$
$\mathrm{R}_{\max }$ is given by $\frac{v_{o}^{2}}{g}$
When $\sin 2 \theta_{0}=1$
For $\mathrm{R}_{\max } \quad \sin 2 \theta_{\mathrm{o}}=1 .$
$\therefore 2 \theta_{\mathrm{o}}=90^{\circ}$
$\therefore \theta_{\mathrm{o}}=45^{\circ}$
If the object is at an angle $\theta_{o}=45^{\circ}$, then its range is maximum. The value of $R_{\max }$ depends upon its initial velocity.
Similar Questions
An object is projected in the air with initial velocity $u$ at an angle $\theta$. The projectile motion is such that the horizontal range $R$, is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be $.......$ degree.
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Given below are two statements. One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion A :Two identical balls $A$ and $B$ thrown with same velocity '$u$ ' at two different angles with horizontal attained the same range $R$. If $A$ and $B$ reached the maximum height $h_{1}$ and $h_{2}$ respectively, then $R =4 \sqrt{ h _{1} h _{2}}$
Reason R: Product of said heights.
$h _{1} h _{2}=\left(\frac{u^{2} \sin ^{2} \theta}{2 g }\right) \cdot\left(\frac{u^{2} \cos ^{2} \theta}{2 g }\right)$
Choose the $CORRECT$ answer
Given below are two statements. One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion A :Two identical balls $A$ and $B$ thrown with same velocity '$u$ ' at two different angles with horizontal attained the same range $R$. If $A$ and $B$ reached the maximum height $h_{1}$ and $h_{2}$ respectively, then $R =4 \sqrt{ h _{1} h _{2}}$
Reason R: Product of said heights.
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Choose the $CORRECT$ answer
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