Question

plzzz give correctly answer​

Answer 1

Given :-

• Velocity of the ball = 5 m/s
• Height of tower = 10 m
• acceleration due to gravity = 10 m/s²

To Find :-

• The Vertical and horizontal components of velocity.
• The velocity of ball

Solution :-

Using third equation of motion ,

$\\ \star \: {\boxed{\purple{\sf{ {v}^{2} - {u}^{2} = 2as}}}}$

Where ,

• v is final velocity
• u is initial velocity
• a is acceleration
• s is distance/height

We have ,

u = 0 m/s [initial velocity along y axis is 0 m/s]

a = g = 10 m/s²

s = h = 10 m

Substituting the values :-

$\\ : \implies \sf \: {v}^{2} - {0}^{2} = 2(10)(10)$

$\\ : \implies \sf \: {v}^{2} = 200$

$\\ : \implies \sf \: v = \sqrt{200}$

$\\ : \implies{\underline{\boxed{\pink{\mathfrak{v = 10 \sqrt{2} \: m {s}^{ - 1} }}}}} \: \bigstar$

So , Vertical component of velocity of ball is $\sf{10\sqrt{2}}$ m/s.

Now We have ,

$\sf{v_y=10\sqrt{2}}$ m/s

vₓ = 5 m/s [Horizontal component remains constant]

Now , Calculating the velocity of ball ,

$\\ : \implies \sf \: v = \sqrt{ {v_x}^{2} + v_y {}^{2} }$

$\\ : \implies \sf \: v = \sqrt{ {(5)}^{2} + {(10 \sqrt{2}) }^{2} }$

$\\ : \implies \sf \: v = \sqrt{25 + 200}$

$\\ : \implies \sf \: v = \sqrt{225}$

$\\ : \implies{\underline{\boxed{\pink{\mathfrak{v = 15 \: m {s}^{ - 1} }}}}} \: \bigstar$

Hence ,

• The Horizontal and vertical components of velocity of given ball are 5 m/s and $\sf{10\sqrt{2}}$ m/s . The velocity of ball is 15 m/s. So , Option(3) is the required answer

Answer 2

$\Huge\mathcal\blue{\underbrace{\overbrace{\mid{\orange{Answer}}\mid}}}$

Given:-

• Height (h) of the tower= 10m
• acceleration due to gravity(g)= 10ms^-1
• Velocity of the ball = 5m/s

To Find:-

• The velocity of the ball when it reaches the ground

Solution:-

For Finding the velocity we will use the Velocity Calculator Formula

$\huge\underline{\overline{\mid{\bold{\blue{\mathcal{v²-u²=2as}}\mid}}}}$

Where,

$v = initial \: velocity$

$u = final \: velocity$

$a = acceeration$

$s = displacement$

We Have,

$u = 0 \frac{m}{s}$

$a = g = 10m {s}^{2}$

$s = h = 10m$

Substituting the values:-

${v}^{2} - {0}^{2} = 2 \times 10 \times 10$

$= > {v}^{2} = 200$

$= > v = \sqrt{200}$

$= > 10 \sqrt{2} m {s}^{ - 1}$

Vertical Component of velocity of ball is 10√2m/s

Now,

$v_{y} = 10 \sqrt{2} \frac{m}{s}$

$v_{x} = 5 \frac{m}{s}$

Now, The Velocity of ball=

$v = \sqrt{v_{x} + v_{y}}$

$= > v = \sqrt{ {5}^{2} + {(10 \sqrt{2}) }^{2} }$

$= > v = \sqrt{25 + 200}$

$= > v = \sqrt{255}$

$= > v = 15m {s}^{ - 1}$

Hence,

The velocity of ball is 15 ms^-1