A cricketer can throw a ball to a maximum horizontal distance of $100\; m$. How much high above (in $m$) the ground can the cricketer throw the same ball ?
Maximum horizontal distance, $R=100 \,m$ The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is $45^{\circ},$ i.e., $\theta=45^{\circ}$ The horizontal range for a projection velocity $v$, is given by the relation
$R=\frac{u^{2} \sin 2 \theta}{g}$
$100=\frac{u^{2}}{g} \sin 90^{\circ}$
$\frac{u^{2}}{g}=100$
The ball will achieve the maximum height when it is throwertically upward. For such motion, the final velocity $v$ is zero at the maximum height $H$ Acceleration, $a=-g$
Using the third equation of motion
$v^{2}-u^{2}=-2 g H$
$H=\frac{1}{2} \times \frac{u^{2}}{g}=\frac{1}{2} \times 100=50 \,m$
Define projectile particle and derive the equation $y\, = \,(\tan \,{\theta _0})x\, - \,\frac{g}{{(2\,\cos \,{\theta _0})}}{x^2}$
A projectile crosses two walls of equal height $H$ symmetrically as shown The height of each wall is ........ $m$
If the range of a gun which fires a shell with muzzle speed $V$ is $R$, then the angle of elevation of the gun is
Given that $u_x=$ horizontal component of initial velocity of a projectile, $u_y=$ vertical component of initial velocity, $R=$ horizontal range, $T=$ time of flight and $H=$ maximum height of projectile. Now match the following two columns.
Column $I$ | Column $II$ |
$(A)$ $u_x$ is doubled, $u_y$ is halved | $(p)$ $H$ will remain unchanged |
$(B)$ $u_y$ is doubled $u_x$ is halved | $(q)$ $R$ will remain unchanged |
$(C)$ $u_x$ and $u_y$ both are doubled | $(r)$ $R$ will become four times |
$(D)$ Only $u_y$ is doubled | $(s)$ $H$ will become four times |
The ceiling of a long hall is $25\; m$ high. What is the maximum horizontal distance that a ball thrown with a speed of $40\; m/ s$ can go without hitting the ceiling of the hall ?