If a stone is to hit a point which is at a ho | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equations of motion for a projectile are:Horizontal displacement: $d = (u \cos 	heta) t$Vertical displacement: $h = (u \sin 	heta) t - \frac

Vedclass · Accepted Answer

$\frac{d}{\cos 	heta} \sqrt{\frac{g}{2(d 	an 	heta - h)}}$

Vedclass · Answer

(B) The equations of motion for a projectile are:Horizontal displacement: $d = (u \cos 	heta) t$Vertical displacement: $h = (u \sin 	heta) t - \frac{1}{2} g t^2$From the horizontal equation,we find the time of flight to reach distance $d$:$t = \frac{d}{u \cos 	heta}$Substitute this expression for $t$ into the vertical displacement equation:$h = (u \sin 	heta) \left( \frac{d}{u \cos 	heta} ight) - \frac{1}{2} g \left( \frac{d}{u \cos 	heta} ight)^2$$h = d 	an 	heta - \frac{g d^2}{2 u^2 \cos^2 	heta}$Rearranging to solve for $u^2$:$\frac{g d^2}{2 u^2 \cos^2 	heta} = d 	an 	heta - h$$u^2 = \frac{g d^2}{2 \cos^2 	heta (d 	an 	heta - h)}$Taking the square root of both sides:$u = \frac{d}{\cos 	heta} \sqrt{\frac{g}{2(d 	an 	heta - h)}}$

Column-$I$	Column-$II$
$i)$ The initial velocity of projection	$a)$ $\sqrt{\frac{g(1+a^2)}{2b}}$
$ii)$ The horizontal range of projectile	$b)$ $\frac{a}{b}$
$iii)$ The maximum height attained by projectile	$c)$ $\frac{a^2}{4b}$
$iv)$ The time of flight of projectile	$d)$ $a\sqrt{\frac{2}{bg}}$

If a stone is to hit a point which is at a horizontal distance $d$ and at a vertical height $h$ above the point from where the stone is launched,what is the initial speed $u$ if the stone is launched at an angle $\theta$?

Similar Questions

Which of the following remains constant for a projectile fired from the earth?

$A$ particle $(A)$ is dropped from a height and another particle $(B)$ is thrown in a horizontal direction with a speed of $5 \; m/s$ from the same height. The correct statement is:

$A$ ball is projected with kinetic energy $E$, at an angle of $60^{\circ}$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will be:

$A$ ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal from a distance $d_1$ from the foot of a pole. It just touches the top of the pole and then falls on the ground at a distance $d_2$ from the pole on the other side. The height of the pole is ...........

Vedclass Test Series

Exam Paper Generator

Online Exam Module