The Equation for the Path of Projectile

Q. a projectile is given an initial velocity of i+2j where i and j are unit vectors along horizontal and vertical directions respectively the equation of its path

Q.

A torque of 20 Nm is applied to a flywheel of mass 10 kg and radius of gyration 50 cm. The resulting angular acceleration is

1. 6 rads^-2

2. 10 rads^-2

3. 8 rads^-2

4. 4 rads^-2

Q. Two blocks of M1 and M2 (m1>m2) are connected by a massless string passing over a frictionless Pulley magnitude of acceleration of block would be. So acceleration would be (m1-m2/m1+m2)×g. So what will the acceleration of m1 which is smaller in wieght

Q. A particle has initial velocity v= 3i +4j . and a a cons†an t force F= 4i -3j acts on the particle. The path of particle is (a)circular (b) parabolic \{I think F perpendicular tov as F.v is zero so it should be circular \}

Q. A body is projected vertically with a speed u .The distance travelled by body in last second of upward journey is same as that of

Q.

The equation of trajectory of a projectile in a vertical planes is given by $y=ax-b{x}^2$ where $a$ and $b$ are constants and, $x$ and $y$ are the horizontal and vertical displacements of the particle from the point of projection. The angle of projection of projectile from the horizontal is

1. ${\tan }^{-1}\left[\frac{1}{b}\right]$

2. ${\tan }^{-1}\left[\frac{a}{b}\right]$

3. ${\tan }^{-1}\left[2\alpha \right]$

4. ${\tan }^{-1}\left[a\right]$

Q. 76.If equations of a projectile projected at angle θ with horizontal are following x = 15t (m) y = 15t 5t2 (m), (where x and y are co-ordinates) then, the speed of ball at maximum height of projectile is

Q. velocity of bullet is 1/n th times initial velocity . No. of plank required to stop the bulle

Q. A projectile is given an initial velocity of $(2\hat{i}+4\hat{j})$$m/s$, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g=10m/s^2$, then the equation of its trajectory is

1. $y=x-5x^2$
2. $y=2x-\frac{5}{4}{x}^2$
3. $y=4x-5x^2$
4. $y=4x-25x^2$

Q. Derive the 3rd Equation of Motion using Calculus Method.

Q. ​an object is thrown vertically upward from the surface of earth . if the upward direction is taken as positive what is the direction of velocity and acceleration of the object during the upward and down ward motion

Q. What are the equations of trajectory motion?

Q. In the given figure, $m_1$ = $m_2$ = $m_3$ = 2 kg. Then, just after cutting string connecting $m_2$ and $m_3$

1. Acceleration of $m_2$ is zero.
2. Acceleartion of $m_3$ is zero.
3. Acceleration of $m_2$ is g/2 upwards.
4. Acceleration of $m_1$ is zero.

Q. 125. A particle is thrown from ground at an angle of 90 with horizontal. During its motion from ground to the highest point, radius of curvature of path (1) Remains constant (2) Decreases gradually (3) Increases gradually (4) First increases then decreases 13

Q. two particles are projected simul†an eously in the same vertical plane from the same point , but with different speeds and at different angles to the horizontal.the path of one particle with respect to other is 1.straight line 2.parabola 3. ellipse 4.circle

Q. A projectile is given an initial velocity of $(2\hat{i}+4\hat{j})$$m/s$, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g=10m/s^2$, then the equation of its trajectory is

1. $y=x-5x^2$
2. $y=2x-\frac{5}{4}{x}^2$
3. $y=4x-5x^2$
4. $y=4x-25x^2$

Q. A projectile is fired from the ground at a speed of 20 m/s at an angle of ${37}^{\circ }$ to the horizontal, then $[g=10m/s^2]$

1. Radius of curvature at firing point is 50 m.
2. As projectile moves, its radius of curvature decreases continuosly.
3. As projectile moves its radius of curvature first decreases and then it increases.
4. Minimum radius of curvature is 30 m.

Q. The path of a projectile is given by $y=ax-b{x}^2$. What is the value of the angle ${\theta }_0$ and speed $v_0$ , about a point at which the projectile is launched?

1. ${\theta }_0={\cos }^{-1}\frac{1}{\sqrt{b^2+1}},v_0=\sqrt{\frac{g}{2b}(1+a^2)}$
2. ${\theta }_0={\cos }^{-1}\frac{1}{\sqrt{a^2+1}},v_0=\sqrt{\frac{g}{2b}(1+a^2)}$
3. ${\theta }_0={\cos }^{-1}\frac{1}{\sqrt{a^2+1}},v_0=\sqrt{\frac{g}{2a}(1+b^2)}$
4. ${\theta }_0={\cos }^{-1}\frac{1}{\sqrt{a^2-1}},v_0=\sqrt{\frac{g}{2b}(1+a^2)}$

Q. Trajectory of particle in a projectile motion is given as $y=x-\frac{x^2}{80}$. Here, $x$ and $y$ are in metre. For this projectile motion match the following and select proper option $(g=10{\text{m/s}}^2)$:
$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(A)}& \text{Angle of projection}& \text{(p)}& 20\text{m}\\ \text{(B)}& \text{Angle of velocity}& \text{(q)}& 80\text{m}\\ & \text{with horizontal}& & \\ & \text{after}4s& & \\ \text{(C)}& \text{Maximum height}& \text{(r)}& {45}^{\circ }\\ \text{(D)}& \text{Horizontal range}& \text{(s)}& {\tan }^{-1}\left(\frac{1}{2}\right)\end{array}$

1. $A\to s,B\to s,C\to q,D\to p$
2. $A\to r,B\to r,C\to q,D\to p$
3. $A\to r,B\to r,C\to p,D\to q$
4. $A\to s,B\to r,C\to p,D\to q$

Q. The equation of motion of a projectile are given by x = 36 t metre and $2y=96t–9.8t^2$ metre. The angle of projection is

1. $si{n}^{-1}\left(\frac{4}{5}\right)$
2. $si{n}^{-1}\left(\frac{3}{5}\right)$
3. $si{n}^{-1}\left(\frac{4}{3}\right)$
4. $si{n}^{-1}\left(\frac{3}{4}\right)$

Q. The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (assuming flat surface) with no surrounding atmosphere are given by $x=6t$ and $y=8t-5t^2$, where $x$ and $y$ are in metre and time $t$ is in second. Find the velocity in $m{s}^{-1}$ with which the body is projected. Take acceleration due to gravity $g=10{\text{m/s}}^2$

Q. 40 the equation of trajectory of a projectile projected with a velocity of 10m/satanangle45degreewithverticalisgivenby(g=10m/s)(a)y^2=10(x+y)(b)y^2=10(x-y)(c)x^2=10(x+y)(d)x^2=10(x-y)

Q. The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y=(8t-5t^2)$ meter and x = 6t meter, where t is in second. The velocity with which the projectile is projected is

1. 8 m/sec
2. 6 m/sec
3. 10 m/sec
4. Not obtainable from the data

Q. The equation of motion of a projectile are given by x = 36 t metre and $2y=96t–9.8t^2$ metre. The angle of projection is

1. $si{n}^{-1}\left(\frac{4}{5}\right)$
2. $si{n}^{-1}\left(\frac{3}{5}\right)$
3. $si{n}^{-1}\left(\frac{4}{3}\right)$
4. $si{n}^{-1}\left(\frac{3}{4}\right)$

Q. Four particles of equal mass are moving round a circle of radius r due to their mutual gravitational attraction. Find the angular velocity of each particle?

Q. A gun is fired horizontally on the bull's eye at a height $h$. Then

1. the bullet hits the bull's eye
2. the bullet moves left or right of the Bull's eye due to jerk experienced on firing
3. the bullet misses the bull's eye and hits upward
4. the bullet misses the target and hits downwards

Q. If a particle is projected from origin and it follows the trajectory $y=x-\frac{1}{2}{x}^2$, then the time of flight is ($g$ = acceleration due to gravity)

1. $\frac{1}{\sqrt{g}}$
2. $\frac{2}{\sqrt{g}}$
3. $\frac{3}{\sqrt{g}}$
4. $\frac{4}{\sqrt{g}}$

Q. A heavy particle is projected with a velocity at an angle with the horizontal into a uniform gravitational field. The slope of the trajectory of the particle varies not according to which of the following curve?

1. 
2. 
3. 
4. 

Q. Find the value of $\theta$ in the diagram, so that the projectile can hit the target, placed at a height of $10m$ from ground (as shown). $(g=10{\text{m/s}}^2$)

1. $\theta ={30}^0$
2. $\theta ={45}^0$
3. $\theta ={\tan }^{-1}5$
4. $\theta ={80}^0$

Q. 10. A particle projected from origin move in x-y plane with a velocity v =3+6xj where and j are the unit vectors along x and y axis. Find the equation of path followed by the particle.