A projectile is given an initial velocity of | vedclass

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

Question

(B) The initial velocity vector is given by $\vec{u} = (1 \hat{i} + 2 \hat{j}) \ ms^{-1}$.Comparing this with $\vec{u} = u_x \hat{i} + u_y \hat{j}$,we

Vedclass · Accepted Answer

$y = 2x - 5x^2$

Vedclass · Answer

(B) The initial velocity vector is given by $\vec{u} = (1 \hat{i} + 2 \hat{j}) \ ms^{-1}$.Comparing this with $\vec{u} = u_x \hat{i} + u_y \hat{j}$,we get the horizontal component $u_x = 1 \ ms^{-1}$ and the vertical component $u_y = 2 \ ms^{-1}$.The equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u_x^2}$.Since $	an 	heta = \frac{u_y}{u_x} = \frac{2}{1} = 2$,we substitute the values into the equation:$y = x(2) - \frac{10 \cdot x^2}{2(1)^2}$$y = 2x - \frac{10x^2}{2}$$y = 2x - 5x^2$.

$A$ projectile is given an initial velocity of $(\hat i + 2\hat j) \ ms^{-1}$,where $\hat i$ is along the ground and $\hat j$ is along the vertical. If $g = 10 \ m/s^2$,the equation of its trajectory is

Similar Questions

$A$ 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 $(g = 10 \, m/s^2)$.........$s$

$A$ ball is projected with a velocity $5 \text{ m/s}$, such that its horizontal range is twice the greatest height attained. The value of the range is (in $\text{ m}$)

The equation of motion of a projectile is $y = 12x - \frac{3}{4}x^2$. The range of the projectile is $..........\,m$.

$A$ particle is projected with a velocity $v$ such that its horizontal range is twice the greatest height attained. The horizontal range is

Two projectiles $A$ and $B$ are thrown with initial velocities of $40\,m/s$ and $60\,m/s$ at angles $30^{\circ}$ and $60^{\circ}$ with the horizontal respectively. The ratio of their ranges is $(g = 10\,m/s^2)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module