What is the solution?Public facilities should be used with great care. ​

Answer:

vertical component of velocity is zero at highest point

so, v= ucos30

v=15× 3½/2

v= 7.5 underoot 3

Given:

Initial speed of the arrow [tex]=15m/s[/tex]

Elevation with respect to horizontal [tex]=30^{o}[/tex]

To find:

Speed at the highest point of the trajectory.

Solution:

Step 1

For a projectile thrown with an initial velocity [tex]u[/tex] at an angle [tex]\alpha[/tex] with respect to the ground, the maximum height reached by the projectile is given by

[tex]H_{max}=\frac{u^{2} .sin^{2}\alpha }{2g}[/tex]

The initial velocity of projection can be resolved into its rectangular components namely [tex]u.cos\alpha[/tex] horizontally and [tex]u.sin\alpha[/tex] vertically.

If we take a look at the vertical component of velocity, this component is responsible for the height achieved by the object.

Just like a ball, which when thrown up, with some initial velocity gets acted upon by the acceleration due to gravity in the downward direction and at the maximum height that is when the potential energy becomes maximum, The Final velocity of the object becomes zero [tex]0m/s[/tex].

Hence,

The value of vertical component of the velocity at the highest point of the projectile will be

[tex]V_{v} =0m/s[/tex]

Step 2

Now,

If we consider the horizontal component of the velocity, [tex]u.cos\alpha[/tex], this component of velocity is responsible for making the object travel the maximum range which is given by

[tex]R=\frac{u^{2}.sin2\alpha }{g}[/tex]

In the horizontal motion of the object, we know that since no acceleration acts on the body during the flight hence, velocity at the beginning of the projectile is exactly equal to the velocity of the projectile at the end of the flight.

Hence, the horizontal component of velocity of the object at the highest point of the projectile is

[tex]V_{h}=u.cos\alpha[/tex]

[tex]V_{h} =15.cos(30^{o} )[/tex]

[tex]V_{h}=15(\frac{\sqrt{3} }{2} )[/tex]

[tex]V_{h}=12.99m/s[/tex]

Step 3

Hence, the velocity of the projectile at the highest point will be

[tex]V=\sqrt{V_{h} ^{2} +V_{v} ^{2} }[/tex]

[tex]V=\sqrt{(12.99)^{2} +(0)^{2} }[/tex]

[tex]V=\sqrt{(12.99)^{2} }[/tex]

[tex]V=12.99m/s[/tex]

Final answer:

Hence, the velocity of the projectile at the highest point of the projectile is 13 m/s (approx).

Answer:

a = 5.61

b = 12.46

c = 45.87

d = 61.16

e = 69.76

f = 141.5

g = 704.92

h = 38.361

Answer:

Given :

Height (H) = 4.9 m

Initial horizontal velocity (u) = 4.5 m/sec

To Find :

Time taken by the stuntman to reach the ground.

Solution :

Velocity of the stuntman in y-axis = ucos∅

                                                        = 4.5×cos90

                                                        = 0

Therefore, applying the Kinematic equation for uniformly accelerated body,

s = ut + [tex]\frac{1}{2} [/tex]gt²

Where, s = distance covered; u = initial velocity; g = acceleration due to gravity; t = time taken

∴    H   = (0)t + [tex]\frac{1}{2} [/tex]gt²

⇒  4.9 =  0  +  [tex]\frac{1}{2} [/tex]×9.8×(t)²

⇒   t²  =  1

⇒   t    = √1

∴    t    = 1 s

The time taken by the stuntman to reach the ground  is 1 sec.

Therefore, option (c) is correct

Grignard's reagent is an organo-magnesium halide having the general chemical formula as (R-Mg-X) which is used to form a Carbon-Carbon bond and reduce an aldehyde or ketone to alcohol.

• "R" is the organic part of the compound. "X" is a halogen, otherly known as group 17 elements like Flourine, Chlorine, etc.
• The compound helps us to create the rather difficult C-C bond or carbon-carbon bond.

• In Grignard's reaction, the desired alkyl, allyl or any carbonyl group is added to the magnesium halide which is then reacted with an aldehyde or ketone to reduce it to alcohol and add the carbonyl group to the organic compound.

The reaction of water on methyl magnesium bromide in dry ether forms methane.

CH₃-Mg-Br + H₂0  --> CH₄ (methane) + Br-Mg-OH

The lone pair electrons on the oxygen atom attack the methyl carbon and remove the MgBr part from the compound and replace it with a Hydrogen atom to form methane.

The action of Zn-HCL on ethyl iodide is that it reduces ethyl iodide to ethane.

CH₃-CH₂-I + 2[H] [tex]\frac{Zn}{HCL}[/tex]-->  CH₃-CH₃ (ethane) +  HI

Zn/HCL acts as a reducing agent thus when reacted with ethyl iodide, the nascent Hydrogen atoms attack the iodine part of the compound and replace it thus forming ethane and Hydroflouride.