Question

Solved the question fully.... please... ​

Answer 1

$\bf{\huge{\underline{\boxed{\mathfrak{\blue{Answer:}}}}}}$

Given:-

Ball is dropped on to the floor from a height of 4 m, $\bf{h_1}$ = 4m

The ball rebounds to a height of 3 m, $\bf{h_2}$= 3m

The ball was in contact with the floor for 0.010s , t = 0.010s

Assumptions :-

Velocity of the ball before when dropped before striking the floor = $\bf{V_1}$

Velocity of the ball after striking the floor = $\bf{V_2}$

As the ball was in contact with the floor for 0.010 sec there must be some changes observed in the velocity of the ball, so,

Change in velocity = $\bf{V_2}$ - (- $\bf{V_1}$ )

Acceleration = $\bf\large\frac{V_2 - (-V_1)}{t}$

Case 1:-

When the ball is dropped from $\bf{h_1}$,

Initial velocity, u = 0

Final velocity, v = $\bf{V_1}$

Acceleration, a = g = 9.8m/s²

Distance, s = $\bf{h_1}$ = 4m

As per the 3 rd equation of motion,

v² = u² + 2as

$\bf{v^2}$ = 0 + 2g$\bf{h_1}$

v² = 0 + 2 × 9.8 × 4

v² = 19.6 × 4

v² = 78.4

v = $bf\squareroot{78.4}$

v = 8.85 m/s

Case 2:-

When the ball rebounds after striking the floor,

Initial velocity = u

u Final velocity, v =0

Acceleration, a = g = 9.8m/s²

Distance, s = $\bf{h_</u><u>2</u><u>}$ =3m

v² = u² + 2as

0 = u² - 2 × 9.8 × 3

0 = u² -19.6 × 3

0 = u² - 58.6

u² - 58.6 = 0

u = √-58.6

u = 7.67m/s

Now calculate the change in velocity,

Change in velocity = 7.67 m/s (upward) – 8.85 m/s (downward)

Δv = ( 7.67 + 8.85 ) m/s

$\bf{\large{\boxed{\blue{<strong> </strong><strong>Δv = 16.52 m/s</strong> }}}}$

Acceleration, a = $\bf\large\frac{Δv}{Δt}$

= $\bf\large\frac{16.52}{0.010}$

= 1652 m/s²

•°• $\bf{\large{\boxed{\red{\mathcal{Acceleration\:due\:to\:contact\:=1652m/s^2}}}}}$