The angle of projection of a particle is meas | vedclass

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

Question

(D) The standard formula for the maximum height $h_m$ of a projectile is given by $h_m = \frac{u^2 \sin^2 	heta}{2g}$,where $	heta$ is the angle of

Vedclass · Accepted Answer

Vedclass · Answer

(D) The standard formula for the maximum height $h_m$ of a projectile is given by $h_m = \frac{u^2 \sin^2 	heta}{2g}$,where $	heta$ is the angle of projection measured from the horizontal axis.Given that the angle of projection $\phi$ is measured from the vertical axis,the angle with the horizontal axis is $	heta = 90^\circ - \phi$.Substituting this into the formula,we get:$h_m = \frac{u^2 \sin^2(90^\circ - \phi)}{2g} = \frac{u^2 \cos^2 \phi}{2g}$.As $\phi$ increases from $0^\circ$ to $90^\circ$,$\cos \phi$ decreases from $1$ to $0$. Therefore,$h_m$ decreases from $\frac{u^2}{2g}$ to $0$ as $\phi$ goes from $0^\circ$ to $90^\circ$. This behavior is represented by the graph in option $D$.

The angle of projection of a particle is measured from the vertical axis as $\phi$ and the maximum height reached by the particle is $h_m$. Here $h_m$ as a function of $\phi$ can be represented as:

Similar Questions

$A$ body of mass $1 \, kg$ starts moving from rest at $t = 0$ in a circular path of radius $8 \, m$. Its kinetic energy varies as a function of time as: $K.E. = 2t^2 \, J$,where $t$ is in seconds. Then:

$A$ projectile has the same range $R$ for two angles of projection. If $T_1$ and $T_2$ are the times of flight in the two cases,then $R$ is

The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is ......... $^o$

Two projectiles are fired from the ground with the same initial speeds from the same point at angles $(45^{\circ}+\alpha)$ and $(45^{\circ}-\alpha)$ with the horizontal direction. The ratio of their times of flight is

Vedclass Test Series

Exam Paper Generator

Online Exam Module