Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of $45^o$ by equal amounts, the ranges are equal”. Prove this statement.
Answer For a projectile launched with velocity
$v _{ o }$ at an angle $\theta_{ o },$ the range is given by
$R=\frac{v_{o}^{2} \sin 2 \theta_{0}}{g}$
Now, for angles, $\left(45^{\circ}+\alpha\right)$ and $\left(45^{\circ}-\alpha\right), 2 \theta_{0}$ is $\left(90^{\circ}+2 \alpha\right)$ and $\left(90^{\circ}-2 \alpha\right),$ respectively. The values of $\sin \left(90^{\circ}+2 \alpha\right)$ and $\sin \left(90^{\circ}-2 \alpha\right)$ are
the same, equal to that of $\cos 2 \alpha .$ Therefore, ranges are equal for elevations which exceed or fall short of $45^{\circ}$ by equal amounts $\alpha$
The $x-t$ graph of a particle moving along a straight line is shown in figure The speed-time graph of the particle is correctly shown by
The velocity at the maximum height of a projectile is half of its initial velocity $u$. Its range on the horizontal plane is
A bullet is dropped from the same height when another bullet is fired horizontally. They will hit the ground
A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3\, ms^{-2}$ for $ 0.5\, minutes$. If the maximum height reached by it is $80\, m$, then the angle of projection is (Take $g = 10\, ms^{-2}$)
A projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ be the times of flights in the two cases, then the product of the two time of flights is proportional to