What is the range of a projectile? Also,provide the velocity of the projectile at its maximum height.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The range $R$ of a projectile is given by the formula: $R = \frac{v_{0}^2 \sin(2\theta_{0})}{g}$,where $v_{0}$ is the initial velocity,$\theta_{0}$ is the angle of projection,and $g$ is the acceleration due to gravity.
At the maximum height,the vertical component of the velocity becomes zero $(v_{y} = 0)$.
Therefore,the velocity of the projectile at the maximum height is equal to its horizontal component: $v = v_{x} = v_{0} \cos \theta_{0}$.

Explore More

Similar Questions

The initial velocity of a projectile is given by $v = a\hat{i} + b\hat{j}$. If the range $R$ is twice the maximum height $H$,then which of the following is true?

Difficult
View Solution

$A$ ball projected at an angle of $45^{\circ}$ with the horizontal crosses two points at equal heights separated by a distance at times $2 \ s$ and $8 \ s$ respectively. The horizontal distance between the two points is (Acceleration due to gravity $= 10 \ m/s^2$) (in $m$)

$A$ ball is projected from the ground with a velocity $V$ at an angle $\theta$ to the vertical. On its path,it makes an elastic collision with a vertical wall and returns to the ground. The total time of flight of the ball is:

Difficult
View Solution

$A$ particle is thrown with a speed of $12 \, m/s$ at an angle $60^o$ with the horizontal. The time interval between the moments when its speed is $10 \, m/s$ is $(g = 10 \, m/s^2)$.........$s$

Difficult
View Solution

The horizontal range $(R)$ of a projectile is $n$ times its maximum height $(H)$. Find the angle of projection $(\theta_0)$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo