**Answer:**

Gamma ray will get to the Earth first due to his frequency of transmission

**Answer:**

**Explanation:**

xf = xita + vxitb + ½axtc.

xf is displacement , dimensional formula L .

Xi initial displacement , dimensional formula L

t is time , dimensional formula T ,

vxi is velocity , dimensional formula LT⁻¹

ax is acceleration , dimensional formula = LT⁻²

xf = xi t a + vxi t b + ½ ax t c.

L = aLT + b LT⁻¹ T + c LT⁻² T

From the law of uniformity , dimensional formula of each term of RHS must be equal to term on LHS

aLT = L

a = T⁻¹

b LT⁻¹ T = L

b = 1 ( constant )

c LT⁻² T = L

c = T

so a = T⁻¹ , b = constant and c = T .

**Answer:**

Speed of bullet = 389.61 m/s.

**Explanation:**

Considering the vertical motion of bullet

Initial vertical speed = 0 m/s

Vertical displacement = 2.9 cm = 0.029 m

Vertical acceleration = 9.81 m/s²

Substituting in s = ut + 0.5at²

0.029 = 0 x t + 0.5 x 9.81 x t²

t = 0.077 s

So ball hits the target after 0.077 s.

Now considering the vertical motion of bullet

Time = 0.077 s

Horizontal displacement = 30 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

30 = u x 0.077 + 0.5 x 0 x 0.077²

u = 389.61 m/s

Speed of bullet = 389.61 m/s.

It is indirectly proportional or inverse so it would be C

The initial velocity in the horizontal direction will be **positive.**

**Answer: Option A**

<u>**Explanation:**</u>

As the ball threw off the building, the ball will follow a projectile path with free fall. So the velocity components will split as horizontal and the vertical velocity. The horizontal components of velocity **remain same **but the vertical velocities vary at rate of 9.8 m/s (approx. 10 m/s) as the ball will be reaching the ground due to free fall.

So gravity will act upon the ball leading to exhibit acceleration due to gravity will falling. Thus with the help of **second equation of motion**, we can determine the time taken by the ball to reach the ground from a height of 150 m. As the second equation of motion is

As the initial velocity of the ball before thrown is zero, u = 0 and as the ball is exhibiting a free fall, so acceleration a will be **equal to acceleration** due to gravity g. And the displacement performed by the ball will equal to height of the building.

Thus,

Thus, the time taken for crossing the cliff is t = 5.5 s.

Now the distance at which the ball falls far from the base of the building is given as 200 m. So the **horizontal velocity **with which the ball is thrown can be found as,

Thus, the horizontal velocity is **positive.**