A ball is thrown from a roof top at an angle of $45^o$ above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have $(a)$ greatest speed $(b)$ smallest speed $(c)$ greatest acceleration - Explain.
A ball is thrown from a roof top at an angle of $45^o$ above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have $(a)$ greatest speed $(b)$ smallest speed $(c)$ greatest acceleration - Explain.
Consider the given figure in which a ball is projected from point 0 and goes to c through $\mathrm{A}$ and $\mathrm{B}$.
Ground
$(a)$ At point B, it will gain the same speed $v_{0}$ and after that speed increases and will be maximum just before reaching at$ C$.
$(b)$ During upward journey from $0$ to $A$ speed decreases and will be minimum at point A.
$(c)$ Acceleration is always constant and is equal to $g$.
Similar Questions
If a stone is to hit at a point which is at a distance $d$ away and at a height $h$ above the point from where the stone starts, then what is the value of initial speed $u$ if the stone is launched at an angle $\theta $ ?
If a stone is to hit at a point which is at a distance $d$ away and at a height $h$ above the point from where the stone starts, then what is the value of initial speed $u$ if the stone is launched at an angle $\theta $ ?
The maximum vertical height to which a man can throw a ball is $136\,m$. The maximum horizontal distance upto which he can throw the same ball is $.....\,m$
The maximum vertical height to which a man can throw a ball is $136\,m$. The maximum horizontal distance upto which he can throw the same ball is $.....\,m$
- [JEE MAIN 2023]
A particle of mass $100\,g$ is projected at time $t =0$ with a speed $20\,ms ^{-1}$ at an angle $45^{\circ}$ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time $t=2\,s$ is found to be $\sqrt{ K }\,kg\,m ^2 / s$. The value of $K$ is $............$ $\left(\right.$ Take $\left.g =10\,ms ^{-2}\right)$
A particle of mass $100\,g$ is projected at time $t =0$ with a speed $20\,ms ^{-1}$ at an angle $45^{\circ}$ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time $t=2\,s$ is found to be $\sqrt{ K }\,kg\,m ^2 / s$. The value of $K$ is $............$ $\left(\right.$ Take $\left.g =10\,ms ^{-2}\right)$
- [JEE MAIN 2023]
A particle is projected with a velocity of $30\,m / s$, at an angle of $\theta_0=\tan ^{-1}\left(\frac{3}{4}\right)$ After $1\,s$, the particle is moving at an angle $\theta$ to the horizontal, where $\tan \theta$ will be equal to $\left(g=10\,m / s ^2\right)$
A particle is projected with a velocity of $30\,m / s$, at an angle of $\theta_0=\tan ^{-1}\left(\frac{3}{4}\right)$ After $1\,s$, the particle is moving at an angle $\theta$ to the horizontal, where $\tan \theta$ will be equal to $\left(g=10\,m / s ^2\right)$
The equation of a projectile is $y =\sqrt{3} x -\frac{ gx ^2}{2}$ the angle of projection is
The equation of a projectile is $y =\sqrt{3} x -\frac{ gx ^2}{2}$ the angle of projection is