$A$ stone is projected at an angle $\theta$ with velocity $u$. If it executes nearly a circular motion at its maximum point for a short time,the radius of the circular path will be ($g=$ acceleration due to gravity).
- A$\frac{u^2}{g}$
- B$\frac{u^2 \cos^2 \theta}{g}$
- C$\frac{u^2 \sin^2 \theta}{g}$
- D$\frac{u^2 \cos^2 \theta}{2g}$
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Similar Questions
$A$ particle is projected at an angle of $60^{\circ}$ with the horizontal from the ground with a velocity $10 \sqrt{3} \ m/s$. The angle between the velocity vector after $2 \ s$ and the initial velocity vector is $(g = 10 \ m/s^2)$. (in $^{\circ}$)
$A$ cricket ball is thrown at a speed of $28 \; m/s$ in a direction $30^{\circ}$ above the horizontal. Calculate:
$(a)$ the maximum height,
$(b)$ the time taken by the ball to return to the same level,and
$(c)$ the distance from the thrower to the point where the ball returns to the same level.
$(a)$ the maximum height,
$(b)$ the time taken by the ball to return to the same level,and
$(c)$ the distance from the thrower to the point where the ball returns to the same level.
Medium
View Solution$A$ body is projected with a speed $u$ at an angle $\theta$ with the horizontal. The radius of curvature of the trajectory,when it makes an angle $\left(\frac{\theta}{2}\right)$ with the horizontal is ($g$ = acceleration due to gravity).
$A$ projectile is thrown with an initial velocity of $(a \hat{i} + b \hat{j}) \text{ m s}^{-1}$. If the range of the projectile is twice the maximum height reached by it,then
Medium
View SolutionThe range of the projectile projected at an angle of $15^{\circ}$ with the horizontal is $50\,m$. If the projectile is projected with the same velocity at an angle of $45^{\circ}$ with the horizontal,then its range will be $........\,m$.
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