Question

an apple falls from a tree and strikes the ground in 0.5s . what is distance covered by the apple ? g = 10 ms-2.​

Answer 1

Answer:

1.25 metres

Explanation:

Given:

Time = 0.5 seconds (t)

Acceleration due to gravity = 10 m/s² (g)

Initial velocity = 0 m/s (u) (As the apple started from rest when falling from the tree)

To find:

Distance covered by apple (s)

Using the second equation of motion, which says:

s=ut+$\frac{1}{2}$at²

Where:

s = distance

u = initial velocity

t = time

a = acceleration

Substituting the values, we get:

s = 0×0.5 + $\frac{1}{2} \times 10 \times 0.5^{2}$

s = 0 + 5×0.25

s = 0+1.25

s = 1.25 metres

The distance covered by the apple is 1.25 metres

Answer 2

Answer:

$\huge\boxed{\fcolorbox{white}{pink}{Answer}}$

Given :-

☞Final Velocity =0m/s

☞Time taken=0.5s

☞Acceleration due to gravity (g) =10m/s square.

Now,

Solution

By 1st equations of Motion

$v = u + at \\ 0 = u + 10 \times 0.5 \\ u = 5ms{}^{ - 1}$

Again,

By using 2nd equations of motion

$v {}^{2} - u {}^{2} = 2gs \\ 0 {}^{2} - 5 { }^{2} = 2 \times 10 \times s \\ - 25 = 20 \times s \\ s = \frac{ - 25}{20} \\ s = - 1.25m \\ s = 1.25m$

distance never be negative so the required answer will be 1.25m

hope this answer helps you