A particle is projected with a velocity $$v$$ so that its range on a horizontal plane is twice the greatest height attained. If $$g$$ is acceleration due to gravity, then its range is:
Since $$R=2H$$
or $$\dfrac{v^2\sin2\theta}{g}=2\times \dfrac{v^2\sin^2\theta}{2g}$$
or $$2 \sin \theta \cos \theta =\sin^2\theta $$ or $$ \tan \theta =2$$
$$R=\dfrac{v^2\sin2\theta }{g}$$
$$=\dfrac{v^2sin\theta \cos \theta }{g}=\dfrac{2v^2}{g}\times \dfrac{2}{\sqrt{5}}\dfrac{1}{\sqrt{5}}=\dfrac{4v^2}{5g}$$
A stone projected with a velocity with a velocity $$u$$ at an angle $$\theta$$ with the horizontal reaches maximum height $$H_{1}$$. When it is projected with the velocity $$u$$ at an angle $$\left (\dfrac {\pi}{2} - \theta\right )$$ with the horizontal, it reaches maximum height $$H_{2}$$. The relation between the horizontal range $$R$$ of the projectile, heights $$H_{1}$$ and $$H_{2}$$ is
A ball is projected from the ground at an angled of $$45^o$$ 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 $$30^o$$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is _________.
A body is thrown from the surface of the earth with velocity 'u' m/s. The maximum height in m above the surface of the earth upto which it will reach is (R = radius of earth, g = acceleration due to gravity)
A particle is projected with a speed u at angle $$\theta$$ with the horizontal. Consider a small part of the trajectory near the highest position and taken it approximately to be circular arc. What is the radius of the circle? This radius is called the radius of curvature of the curve at the point.
A particle is projected with a speed u at angle $$\theta$$ with the horizontal. Consider a small part of the trajectory near the highest position and taken it approximately to be a circular arc. What is the curvature of the parabola traced out by the projectile at a point where the particle velocity makes an angle $$\theta /2$$ with the horizontal?
Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed u at an angle $$\theta $$ with the horizontal.
A ball is projected vertically upward with a speed of $$50\ m/s$$. Find (a) the maximum height, (b) the time to reach the maximum height (c) the speed at half the maximum height. Take $$g=10\ m/s^{2}$$.
A ball is thrown at a speed of $$40\ m/s$$ at an angle of $${60^0}$$ with the horizontal. Find (a) the maximum height reached and (b) the range of the ball. Take $$g = 10\ m/{s^2}$$.
A ball of mass M is thrown vertically upwards. Another ball of mass 2M is thrown at an angle $$\theta$$ with the vertical. Both of them stay in air for the same period of time. The heights attained by the two are in the ratio.
Two stones having different masses $$m_1$$ and $$m_2$$ are projected at an angle $$\alpha$$ and $$(90^o-\alpha)$$ with same velocity from same point. The ratio of their maximum heights is?