Question

A ball is thrown from the ground level so as to just clear a wall 4 m high at a distance of 4m and falls at a distance of 14 m from the wall.Find magnitude and direction of velocity of the ball??

Answer 1

-4.9 t^2 +A sin(Vi) t = 0

A cos(Vi) t = 28

would be equations to work with i think

Answer 2

t = 28/A cos(Vi)

-4.9 A sec(Vi)^2 +28 Atan(Vi) = 0

maybe? but i gotta get to class ..... good luck with it :)

Answer 3

2.A particle is projected with a velocity v=ai+bj.Find the radius of curvature of trajectory of particle at the
(1)point of projection
(2)highest point

Answer 4

\[v = ai + b j\]
hmm Aravind a, b are constants right...but theres a problem the y velocity should depend on x or else when I differentiate it I get zero which implies no acceleration while in projectile motion the acceleration on y axis is -g...

Answer 5

hmm if you want to know the Radius for general projectile, I can get that ...wait a minute let me post the solution

Answer 6

i need the answer without using differentiation

Answer 7

Let \(v_i\) be the speed with which projectile is projected.
Then at Highest Point \(v_y=0\) And \(v_x\) Remains constant as there is no acceleration on x axis or horizontal component of velocity.
\(\theta\) be the angle at which projectile is projected.
\[v_icos\theta=v_{x}\]
hmm Now....\(\frac{v^2}{R}=a\)
\[R = \frac{v^2}{a}\]
\[a=g\]\[v=v_x=v_i*cos\theta\]\[R=\frac{v_i^2*cos^2\theta}{g}\]

Answer 8

The Answer I posted is for Radius at Highest Point ....

Answer 9

k R= radius and not range ryt??

Answer 10

Yea..R is Radius

Answer 11

nikvist???

Answer 12

...........

Answer 13

........................................

Answer 14

\[y=y(x)\quad;\quad A(0,0)\quad,\quad B(4,4)\quad,\quad C(18,0)\]
\[y=px^2+qx+r\]
\[A(0,0)\quad:\quad 0=r\]
\[y=x(px+q)\quad\Rightarrow\quad 4=4(4p+q)\quad,\quad 0=18(18p+q)\quad\Rightarrow\quad q=-18p\]
\[4p+q=1=4p-18p\quad\Rightarrow\quad p=-\frac{1}{14}\quad,\quad q=\frac{18}{14}\quad\Rightarrow\quad y=-\frac{1}{14}x^2+\frac{18}{14}x\]
\[y'=-\frac{2}{14}x+\frac{18}{14}\quad,\quad\tan\theta=y'(0)=\frac{9}{7}\quad\Rightarrow\quad\theta=\arctan\frac{9}{7}=52.125^{o}\]
\[d_\max=\frac{v^2_0}{g}\sin{2\theta}\quad,\quad v_0=\sqrt{\frac{gd_\max}{\sin{2\theta}}}=13.63\,\,m/s\quad(g\approx10m/s^2)\]

Answer 15

math+physics method

Answer 16

Cool Method .... Thanks I've never seen anyone doing this before .... : )..

Answer 17

Method is Super COOL.......

Answer 18

can u explain wat u did??

Answer 19

ishaan help me here
http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e6649b40b8b1f45b4b73f63

Answer 20

I find equation of trajectory y(x)

Answer 21

what is the answer of first question???i am not clear what amistre said

Answer 22

answer is in my first reply

Answer 23

k now 2nd question

Answer 24

3.Two cars A and B start off to a race on a straight path with initial velocities 8m/s and 5 m/s respectively.Car A moves with uniform acceleration of 1 m/s^2 and car B moves with uniform acceleration 1.1m/s^2.If both the cars reach the winning post together,find length of the race track.Also find which of the two cars was ahead 10 s before the finish.

Answer 25

\[\vec{v}=a\vec{i}+b\vec{j}\quad,\quad\tan\theta=\frac{b}{a}\]
(1) point of projection \[R_1=\frac{v^2_1}{a_{1\perp}}=\frac{a^2+b^2}{g\cdot\cos\theta}=\frac{a^2+b^2}{g}\sqrt{1+\tan^2\theta}=\frac{a^2+b^2}{g}\sqrt{1+\frac{b^2}{a^2}}=\frac{\left(a^2+b^2\right)^{3/2}}{ga}\]
(2) highest point
\[R_1=\frac{v^2_2}{a_{2\perp}}=\frac{a^2}{g}\]

Answer 26

ishaan got the answeer???

Answer 27

umm No..I am on it ...

Answer 28

nikvist???

Answer 29

wow nikvist plz help in 3rd question

Answer 30

4.A car accelerates from rest with a constant acceleration \[\alpha\] on a straight road.After gaining a velocity v,the car moves with that velocity for sometime.Then the car decelerates with a retardation \[\beta\].If total distance covered by car is equal to s ,find total time of its motion.

Answer 31

\[s_A=v_{0A}t+\frac{1}{2}a_At^2\quad,\quad s_B=v_{0B}t+\frac{1}{2}a_Bt^2\]
\[s_A=s_B\quad\Rightarrow\quad v_{0A}t+\frac{1}{2}a_At^2=v_{0B}t+\frac{1}{2}a_Bt^2\quad\Rightarrow\quad (v_{0A}-v_{0B})t=\frac{1}{2}(a_B-a_A)t^2\]
\[t=2\frac{v_{0A}-v_{0B}}{a_B-a_A}=2\frac{3\,\,m/s}{0.1\,\,m/s^2}=60s\]
\[s=8\cdot 60+\frac{1}{2}\cdot 3600=2280\,\,m\]
A was ahead, because it started fast

Answer 32

i became a big fan of u .Now 4th one

Answer 33

5.A racing motor speeds up in a straight line in a lake ,from rest .Referring to the accleration -displacement graph for the speeding boat,find its speed when it passes a raft at a distance of s from starting point.

Answer 34

\[s=s_1+s_2+s_3\]
\[s_1=\frac{v^2}{2\alpha}\quad,\quad t_1=\frac{v}{\alpha}\]
\[s_3=\frac{v^2}{2\beta}\quad,\quad t_3=\frac{v}{\beta}\]
\[s_2=s-s_1-s_3=s-\frac{v^2}{2\alpha}-\frac{v^2}{2\beta}\quad\Rightarrow\quad t_2=\frac{s_2}{v}=\frac{s}{v}-\frac{v}{2\alpha}-\frac{v}{2\beta}\]
\[t=t_1+t_2+t_3=\frac{v}{\alpha}+\frac{s}{v}-\frac{v}{2\alpha}-\frac{v}{2\beta}+\frac{v}{\beta}=\frac{s}{v}+\frac{v}{2\alpha}+\frac{v}{2\beta}\]

Answer 35

you rock SUPER COOL now 5th one

Answer 36

nikviiiiiiiiiiii

Answer 37

\[s=\frac{v^2}{2\alpha}+\frac{v^2}{2\beta}=v^2\frac{\alpha+\beta}{2\alpha\beta}\]
\[v=\sqrt{\frac{2\alpha\beta}{\alpha+\beta}\,s}\]

Answer 38

6.Prove that path of a projectile with respect to another projectile is a straight line...

Answer 39

thx now 6

Answer 40

7.Two bodies are projected simultaneously with mutually perpendicular velocities each of magnitude v0=\[\sqrt{2*g*h}\] from top and bottom of a cliff of height h=8m.If the bodies collide in air,how far from the foot of the cliff does the collision occur?

Answer 41

\[\vec{u}(t)=u_{0x}\vec{i}+(u_{0y}-gt)\vec{j}\]
\[\vec{v}(t)=v_{0x}\vec{i}+(v_{0y}-gt)\vec{j}\]
\[\vec{v}_{rel}=\vec{v}-\vec{u}=v_{0x}\vec{i}+(v_{0y}-gt)\vec{j}-u_{0x}\vec{i}-(u_{0y}-gt)\vec{j}=(v_{0x}-u_{0x})\vec{i}+(v_{0y}-u_{0y})\vec{j}\]
\[\vec{v}_{rel}\neq\vec{v}_{rel}(t)\] one direction=straight line

Answer 42

wow now 7th

Answer 43

8.A body is thrown at an angle theta with horizontal such that it attains a speed equal to \[\sqrt{2/3}\] times the speed of projection when the body is at half of its maximum height.Find the angle theta.

Answer 44

pause after 7th