The equations of motion of a projectile are g | vedclass

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

Question

(A) The horizontal position is given by $x = 36t$. The horizontal velocity is $v_x = \frac{dx}{dt} = 36 \, m/s$.The vertical position is given by $2y

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{4}{5}\right)$

Vedclass · Answer

(A) The horizontal position is given by $x = 36t$. The horizontal velocity is $v_x = \frac{dx}{dt} = 36 \, m/s$.The vertical position is given by $2y = 96t - 9.8t^2$,which simplifies to $y = 48t - 4.9t^2$. The vertical velocity is $v_y = \frac{dy}{dt} = 48 - 9.8t$.At the time of projection $(t = 0)$,the initial horizontal velocity is $u_x = 36 \, m/s$ and the initial vertical velocity is $u_y = 48 \, m/s$.The angle of projection $	heta$ is given by $	an 	heta = \frac{u_y}{u_x} = \frac{48}{36} = \frac{4}{3}$.Using the trigonometric identity for a right-angled triangle with opposite side $4$ and adjacent side $3$,the hypotenuse is $\sqrt{4^2 + 3^2} = 5$. Therefore,$\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{4}{5}$.Thus,$	heta = \sin^{-1}\left(\frac{4}{5}ight)$.

The equations of motion of a projectile are given by $x = 36t \, m$ and $2y = 96t - 9.8t^2 \, m$. The angle of projection is

Similar Questions

In the graph shown in the figure, which quantity associated with projectile motion is plotted along the $y$-axis?

$A$ projectile thrown with a speed $v$ at an angle $\theta$ has a range $R$ on the surface of the Earth. For the same $v$ and $\theta$,its range on the surface of the Moon will be:

$A$ particle of mass $m$ is projected with a velocity $u$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is:

$A$ cricketer can throw a ball to a maximum horizontal distance of $100\, m$. The speed with which he throws the ball is ......... $m/s$ (to the nearest integer).

Two objects are projected with same velocity '$u$' at different angles $\alpha$ and $\beta$ with the horizontal. If $\alpha + \beta = 90^{\circ}$, the ratio of horizontal range of the first object to the second object will be $[\sin(\pi - \theta) = \sin \theta]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module