Question

A ball starts from rest and rolls down 16 m down an incillined plane is 4 s what is the velocity of the ball at the bottom of the incline

Answer 1

Given:-

• Initial velocity ,u = 0m/s (at rest)

• Distance ,s = 16

• Time taken ,t = 4s

To Find:-

• Final velocity ,v ?

Solution:-

⠀⠀ According to the Question

As the ball start roll down from an inclined plane . So, we calculate the acceleration of the ball. Using 2nd equation of motion.

• s = ut + 1/2 at²

substitute the value we get

$:\implies$ 16 = 0×4 + 1/2×a × 4²

$:\implies$ 16 = 0 + 1/2 × 16 × a

$:\implies$ 16 = 8a

$:\implies$ a = 16/8

$:\implies$ a = 2 m/s²

Now using 3rd equation of motion

• v² = u² + 2as

where

• v denote final velocity
• u denote initial velocity
• a denote acceleration due to gravity
• s denote distance covered

Substitute the value we get

$:\implies$ v² = 0² + 2 × 2×16

$:\implies$ v² = 0 + 4 × 16

$:\implies$ v² = 64

$:\implies$ v = √64

$:\implies$ v = 8 m/s

• Hence, the final velocity of the ball is 8 m/s .

Answer 2

Given :-

• A ball starts from rest and rolls down 16 m down an inclined plane is 4 s.

To Find :-

• What is the velocity of the ball at the bottom of the incline.

Formula Used :-

$\clubsuit$ Second Equation Of Motion Formula :

$\longmapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2}at^2}}}$

$\clubsuit$ Third Equation Of Motion Formula :

$\longmapsto \sf\boxed{\bold{\pink{v^2 - u^2 =\: 2as}}}$

where,

• s = Distance Covered
• u = Initial Velocity
• t = Time
• a = Acceleration
• v = Final Velocity

Solution :-

First, we have to find the acceleration :

Given :

• Distance Covered = 16 m
• Initial Velocity = 0 m/s
• Time = 4 seconds

According to the question by using the formula we get,

$\implies \sf 16 =\: 0(4) + \dfrac{1}{2} \times a(4)^2$

$\implies \sf 16 =\: 0 + \dfrac{1}{2} \times a(16)$

$\implies \sf 16 - 0 =\: \dfrac{1}{2} \times 16a$

$\implies \sf 16 =\: \dfrac{1}{2} \times 16a$

$\implies \sf 16 \times 2 =\: 16a$

$\implies \sf 32 =\: 16a$

$\implies \sf \dfrac{\cancel{32}}{\cancel{16}} =\: a$

$\implies \sf 2 =\: a$

$\implies \sf \bold{\purple{a =\: 2\: m/s^2}}$

Hence, the acceleration is 2 m/s².

Now, we have to find the final velocity :

Given :

• Initial Velocity = 0 m/s
• Acceleration = 2 m/s²
• Distance Covered = 16 m

According to the question by using the formula we get,

$\longrightarrow \sf (v)^2 - (0)^2 =\: 2(2)(16)$

$\longrightarrow \sf (v)^2 - 0 =\: 2 \times 2 \times 16$

$\longrightarrow \sf (v)^2 =\: 4 \times 16$

$\longrightarrow \sf (v)^2 =\: 64$

$\longrightarrow \sf v =\: \sqrt{64}$

$\longrightarrow \sf\bold{\red{v =\: 8\: m/s}}$

$\therefore$ The velocity of the ball at the bottom of the incline is 8 m/s.