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:

Apple got dropped from the branch of a tree and hits the ground in 0.5 seconds. Acceleration of the Apple during its motion is 10 m/s².

To find:

• Speed just before it hits the ground

• Average speed during 0.5 seconds

• Height of the branch of the tree.

Calculation:

We shall apply the equations of kinematics because the gravitational acceleration is assumed to be constant.

Applying 1st Equation of Kinematics:

$v = u + at$

$= > v = 0 + (10 \times 0.5)$

$= > v = 5 \: m {s}^{ - 1}$

So velocity just before hitting ground is 5 m/s.

Applying 3rd Equation of Kinematics:

${v}^{2} = {u}^{2} + 2gh$

$= > {5}^{2} = {0}^{2} + (2 \times 10 \times h)$

$= > 20h = 25$

$= > h = \dfrac{25}{20}$

$= > h = 1.25 \: m$

So the branch of the tree is located 1.25 metres above the ground.

Now, average velocity is defined as the ratio of of total displacement to the total time taken.

$v \: avg. = \dfrac{total \: displacement}{total \: time}$

$= > v \: avg. = \dfrac{1.25}{0.5}$

$= > v \: avg. = 2.5 \: m {s}^{ - 1}$

So, average velocity is 2.5 m/s

Answer 2

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 m/s² in the downward direction, find:-

• Speed just before it hits the ground
• Height of branch of tree from the ground
• Average velocity during whole journey

Sign Convention:-

• Acceleration = -g
• Velocity while hitting = negative
• Initial velocity = 0 m/s [Dropped means released from rest]
• Displacement = negative

Answer:-

1st answer:-

It is already given to us that the time of flight is 0.5 seconds

Apply first equation of motion:-

v = u + at

➵ v = 0 - 10 * 0.5

➵ v = - 5 m/s

Velocity is a vector quantity and -ve sign indicates downward direction.

Hence, speed = | v | = | -5 | = 5 m/s

Speed while hitting = 5m/s

2nd answer:-

Apply second equation of motion:-

s = ut + 1/2 * a * t²

➵s = 0 - (1/2)(10)(0.5)²

➵s = - 5 * 0.25

➵s = - 1.25 m

Displacement is a vector quantity and -ve sign indicates downward direction

Height means length(or distance) which is | s |

Therefore, height of tree = | - 1.25 | = 1.25m

Height of branch of tree from ground = 1.25m

3rd answer:-

Average velocity = (Total Displacement)/(Total time)

➵ Here, Average velocity during 0.5 seconds = (-1.25)/(0.5) = -2.5 m/s

Velocity is a vector quantity and -ve sign indicates downward direction.

Average velocity during 0.5 seconds = -2.5 m/s