1. Add the following numbers. a) 3.45 + 2.16 b) 4.56 + 7.9 c) 10.90 + 34.97 d) 32.11 + 29.05 e) 11.98 + 34.56 + 23.22 f) 25.456 + 3.654 + 112.39g) 6.52 + 63.2 + 635.2 h) 12.45 + 24.67 + 1.241​

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that, height of cylinder exceeds the radius by 7 cm

Let assume that

Radius of cylinder = r cm

Height of cylinder = h = r + 7

According to statement, it is further given that the curved surface area of a cylinder is three times the base area.

We know,

Curved Surface Area of cylinder of radius r and height h is given by

[tex]\boxed{ \rm{ \:CSA_{(Cylinder)} \: = \: 2 \: \pi \: r \: h \: \: }} \\ [/tex]

and

[tex]\boxed{ \rm{ \:Base \: Area_{(Cylinder)} \: = \: \pi \: {r}^{2} \: \: }} \\ [/tex]

So,

[tex]\rm \: 2\pi \: rh \: = \: 3\pi {r}^{2} \\ [/tex]

[tex]\rm \: 2h \: = \: 3r \\ [/tex]

On substituting the value of h, we get

[tex]\rm \: 2 \:(r + 7) \: = \: 3r\\ [/tex]

[tex]\rm \: 2r + 14 \: = \: 3r\\ [/tex]

[tex]\rm \: 3r - 2r = 14\\ [/tex]

[tex]\rm\implies \:r \: = \: 14 \: cm \\ [/tex]

So, we have

Radius of cylinder, r = 14 cm

Height of cylinder, h = r + 7 = 14 + 7 = 21 cm

Now,

[tex]\rm \: Volume _{(Cylinder)} \: = \: \pi \: {r}^{2} \: h \\ [/tex]

[tex]\rm \: =  \:\dfrac{22}{7} \times 14 \times 14 \times 21 \\ [/tex]

[tex]\rm \: =  \:22 \times 2 \times 14 \times 21 \\ [/tex]

[tex]\rm \: =  \:12936 \: {cm}^{3} \\ [/tex]

Hence,

[tex]\color{green}\rm\implies \:\boxed{ \rm{ \:Volume _{(Cylinder)} = 12936 \: {cm}^{3} \: }} \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]

Given:

The height of the cylinder exceeds the radius by 7cm.

The CSA of the cylinder is 3 times the base area.

To Find:

The volume of the cylinder

Solution:

Let the radius of the cylinder be r

Height of the cylinder h = r + 7

It is given that the curved surface area of the cylinder is three times the base area.

The formula to calculate the curved surface area of the cylinder is

CSA of the cylinder = 2πrh

The base area of the cylinder can be calculated by using πr²

Now, substituting the values

2πrh = 3πr²   [ it is given that 3 times the base area]

2h = 3r

Then we put the value of h we get

⇒ 2(r+7) = 3r

⇒ 2r + 14 = 3r

⇒ 3r - 2r = 14

⇒ r = 14cm

∴ The radius of the cylinder is 14cm.

The height of the cylinder h = r + 7

                                            h = 14 + 7

                                             h = 21cm

∴ The height of the cylinder = 21cm

Now, Volume of the cylinder = πr²h

                                                = 22/7 × 14 × 14 × 21

                                                = 22 × 2 × 14 × 21

                                                = 12936cm³

Therefore the volume of the cylinder = 12936cm³.

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:

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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