A ball is projected from ground at an angle of 45^{circ} with horizontal surface. It reaches a maxim

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3-2.Motion in Plane

normal

A ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal surface. It reaches a maximum height of $120 m$ and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of $30^{\circ}$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is. . . . .

A$20$

B$30$

C$40$

D$50$

(IIT-2018)

Solution

$H =\frac{ u ^2 \sin ^2 45^{\circ}}{2 g }$
$h =\frac{ v ^2 \sin ^2 30^{\circ}}{2 g }$
$\frac{ h }{ H }=\frac{ v ^2 \times \frac{1}{4}}{ u ^2 \times \frac{1}{2}}=\frac{1}{2}$$\left(\frac{ v ^2}{ u ^2}\right)=\frac{1}{4}$

Standard 11

Physics

A projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$, the range of the projectile is $d$. If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is $g^{\prime}=\frac{g}{0.81}$, then the new range is $d^{\prime}=n d$. The value of $n$ is. . . . .

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(IIT-2022)

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(AIPMT-1997)

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hard

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Solution

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