Hello,

The velocity of ball is u m/s

The velocity of descending balloon is v m/s

Therefore, the velocity of ball w.r.t. balloon is u + v m/s

Now, the ball will pass the balloon when the relative displacement between the two is zero.

Now, displacement is given by, s = u't + 1/2 * a * t^2

= s = (u + v)*t - 1/2 * g * t^2 = 0 [ a = (-g) = acceleration due to gravity and u' = u + v]

Taking t common, we have:

t * [ (u + v) - 1/2 * g * t ] = 0 which gives t = 0 or [ (u + v) - 1/2 * g * t ] = 0

Since t cannot be equal to 0, therefore, we have: [ (u + v) - 1/2 * g * t ] = 0

Rearranging and solving the above equation, we get:

t = 2 * ( u + v ) / g (Answer)

Hello,

The displacement of a body is given by s = ut + 1/2 * a * t^2

For a freely falling body, u = 0

Therefore, s1 = 1/2 * a * t^2 = 1/2 * 10 * t^2 = 5 * t^2

Therefore, s1 = 5 * t ^2

For a body falling with an initial velocity of 2 m/s, displacement after time t second will be, s2:

s2 = 2 * t + 1/2 *10*t^2

s2 = 2t + 5 * t^2

Now, when the separation between them is 18 m ( I have assumed 18 m in place of 18 years because I think you have typed it incorrect)

Therefore, s2 - s1 = 18

2t + 5 * t^2 - 5 * t^2 = 18

2t = 18

t = 9 seconds (Answer)

Hello,

Assume the length of the spring to be 'l'

Now the ratios of lengths in which the spring is cut is 1:2:3

That means the length of the three new springs will be x, 2x and 3x where x is equal to some constant.

Therefore, 1x + 2x + 3x = l

= 6x = l

= x = l/6

Now length of new springs will be l/6, 2l/6 and 3l/6

(Substituting the value of x = l/6)

Hope it is clear now.

Draw this answer on x-axis and y-axis for better understanding.

Let your tent is at a point, we say it O (0,0), Joe's tent is at Point J (in the first quadrant) and Karl's tent is at Point K (in the fourth quadrant). So now acc. to the ques. line joining OJ will make angle of 37.0 and line joining OK will make angle of 23.0 with the x-axis.

Given, OJ = 21m. and OK = 32m.

Therefore; vector OJ = (21xcos23)i - (21xsin23)j

and vector OK = (32xcos37)i + (32xsin37)j

From vector triangle,

OJ + JK = OK => JK = OK - OJ

vector JK = (32xcos37 - 21xcos23)i + (32xsin37 + 21xsin23)j

|JK|^2 = [(32xcos37 - 21xcos23)^2 + (32xsin37 + 21xsin23)^2] m^2

Jk         = 28.16m

So, the distance between Joe and Karl tent is = 28.16m or 28.20m

Hello

The equation g÷ 2u^2cos^2theta is used to solve Projectile Motion Problems to make make that equation represents Equation of Parabola

Check the Derivation here

Let projectile is thrown with a initial velocity u at an angle theta there with horizontal.

U=uxi+yuk

Ux=ucos theta

Uy=usin theta

Neglecting air resistance

Along x direction

S=ut+1/2at^2

Distance covered by projectile = velocity time

t=X(x)/ucos theta ----------(1)

Along y direction

S=ut+1/2at^2

y=usin theta × t-1/2gt^2 ---------(2)

Substituting (1) in (2)

y= usin theta × x/ucos theta - 1/2 g (x/ucos theta)^2

y=x tan theta - (g/2u^2cos^2theta

Let tan theta =A

-g/2u^2cos^2 theta =B

Y= Ax+Bx^2--------(Parabola)

This equation represents equation of parabola.

Hence path of projectile(Trajectory) is a parabola

Hope this Helps

All the best

There are seven physical fundamental quantities. Its been listed in the Measurement chapter of Physics 11th standard book as well.

I will enlist them down below with their SI units..

1. Length (m)
2. Mass (kg)
3. Time (s)
4. Electric current (amp)
5. temperature (kelvin)
6. amount of a substance (mol)
7. luminous intensity (candela)

Best wishes. Thank you.