Question

An apple dropped from the branch of a tree hits the ground in 0.5 s. If the acceleration of the apple during its motion is 10 ms-2 in the downward direction. Calculate ----- Its speed just before it hits the ground. What is its average velocity during 0.5 s? Calculate the height of the branch of the tree from the ground?

Answer 1

Given that,

Time = 0.5 s

Acceleration = 10 m/s²

(I). We need to calculate the speed of apple

Using equation of motion

`v=u+at`

Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

`v=0+10*0.5`

`v=5\ m/s`

(III). We need to calculate the height of the branch of the tree from the ground

Using equation of motion

`s=ut+(1)/(2)gt^2`

Put the value into the formula

`s=0+(1)/(2)*10*(0.5)^2`

`s=1.25\ m`

(II). We need to calculate the average velocity during 0.5 sec

Using formula of average velocity

`v_(avg)=(\Delta x)/(\Delta t)`

`v_(avg)=(x_(f)-x_(i))/(t_(f)-t_(0))`

Where,
`x_(f)`= final position

`x_(i)` = initial position

Put the value into the formula

`v_(avg)=(1.25+0)/(0.5)`

`v_(avg)=2.5\ m/s`

Hence, (I). The speed of apple is 5 m/s.

(II). The average velocity during 0.5 sec is 2.5 m/s

(III). The height of the branch of the tree from the ground is 1.25 m.