Suppose a player hits several baseballs. Which baseball will be in the air for the longest time?
The one with the farthest range.
The one which reaches maximum height.
The one with the greatest initial velocity.
The one leaving the bat at $45^o$ with respect to the ground.
A projectile is launched from the origin in the $xy$ plane ( $x$ is the horizontal and $y$ is the vertically up direction) making an angle $\alpha$ from the $x$-axis. If its distance. $r =\sqrt{ x ^2+ y ^2}$ from the origin is plotted against $x$, the resulting curves show different behaviours for launch angles $\alpha_1$ and $\alpha_2$ as shown in the figure below. For $\alpha_1, r ( x )$ keeps increasing with $x$ while for $\alpha_2$, $r(x)$ increases and reaches a maximum, then decreases and goes through a minimum before increasing again. The switch between these two cases takes place at an angle $\alpha_c\left(\alpha_1 < \alpha_c < \alpha_2\right)$. The value of $\alpha_c$ is [ignore where $v_0$ is the initial speed of the projectile and $g$ is the acceleration due to gravity]
Range of a bob of mass $1\, kg$ is $80\,m$ while it attains maximum height $20\,m$. Its change in momentum in complete motion ............ $N-s$
A ball of mass $1 \;kg$ is thrown vertically upwards and returns to the ground after $3\; seconds$. Another ball, thrown at $60^{\circ}$ with vertical also stays in air for the same time before it touches the ground. The ratio of the two heights are
Two projectiles $A$ and $B$ are thrown with the same speed but angles are $40^{\circ}$ and $50^{\circ}$ with the horizontal. Then
A boy playing on the roof of a $10\, m$ high building throws a ball with a speed of $10\, m/s$ at an angle $30^o$ with the horizontal. ........ $m$ far from the throwing point will the ball be at the height of $10\, m$ from the ground . $(g \,= \,10 m/s^2, \,sin \,30^o \,= \,\frac{1}{2}$, $\cos \,{30^o}\, = \,\frac{{\sqrt 3 }}{2}$)