Question

A ball is projected horizontally with a velocity of 5ms from the top of a tower of
height 10m. When it reaches the ground
(g=10ms-2)
1) vertical component of its velocity is 15ms-1
2) Horizontal component of its velocity is 14ms-1
3) Its velocity is 15ms-1
4) Its velocity is root29 ms​

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

3) Its velocity is 15ms-1

hope this answers will help you