Question

solve it................ ​

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

☘️ See the attachment picture for Explanation.

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