Question

A balloon rises with constant acceleration of 1.25 metre per?​ second square from ground after eight second a stone is dropped from the balloon then find out
A. Displacement of the stone till it reaches on the ground
B. Total distance travelled by the stone​

Answer 1

Answer:

A. 40m

B. 50m

Explanation:

Given

Acceleration is constant = 1.25 m/s^

Total Time = 8 sec

To Find

A. Displacement of the stone till it reaches on the ground

B. Total distance travelled by the stone

Solutions;

A. For Displacement

According to Question

After 8 seconds

Height of the balloon = Displacement

of the stone

$h = \frac{1}{2} a {t}^{2} \\ = > h = \frac{1}{2} \times 1.25 \times ( {8}^{2} ) \\ = > h = \frac{1}{2} \times 1.25 \times 8 \times 8 \\ = > h = \frac{80}{2} \\ = > \huge \boxed{h = 40m}$

B. For total time travelled by the stone

On using 1st equation of motion

$= > v = u + at \\ = > v = 0 \times 1.25 \times 8 \\ = > \boxed{v = 10}$

Now

We know that

$= > {v}^{2} = {u}^{2} - 2gx \\ = > 0 = {10}^{2} - 2 \times 10 \times x \\ = > x = \frac{ \cancel100}{ \cancel20} \\ = > \boxed{x = 5}$

$Total \: Distance \: Travelled \\ = 5 + 5 + 40 \\ \huge \boxed{Total \: Distance = 50m}$

where,

a = acceleration

u = initial velocity

v = final velocity

x = height

Answer 2

Explanation:

Displacement = 1/2 * 1.25 * 8 * 8

Displacement = 80/2

Displacement = 40 m

Total distance = 40 + 5 +5

Total distance = 50m