Question

An orange drops to the ground from the top of a tree 45m tall. How long does it take to reach the ground? (g = 10m/s2

Answer 1

Given:-

• Initial Velocity = 0m/s.

• Height of the tree = 45m

• Acceleration due to gravity = +10m/s²

To Find:-

The Time taken by orange to reach ground.

FORMULAE USED:-

• h = ut + ½ × g × t²

Where,

h = Height

u = Initial Velocity

g = Acceleration due to gravity

t = time

Now,

h = ut + ½ × g × t²

h = 0 × t + ½ × 10 × t²

45 = 0 + 5t²

45 = 5t²

t² = 45/5

t² = 9

√t² = √9

t = 3s.

Hence, The Time taken by orange to fall is 3 Second.

Answer 2

$\orange{\bigstar}$ Answer $\green{\bigstar}$

Initial velocity , u = 0 m/s

∵ Drops from top of a tree

Height , s = 45 m

Acceleration due to gravity , a = 10 m/s²

Time , t = ? s

Apply 2nd equation of motion ,

$:\to\ \bf s=ut+\dfrac{1}{2}at^2\ \; \orange{\bigstar}\\\\:\to \rm (45)=(0)t+\dfrac{1}{2}(10)t^2\\\\:\to \rm 45=0+5t^2\\\\:\to \rm 45=5t^2\\\\:\to \rm 5t^2=45\\\\:\to \rm t^2=9\\\\:\to \rm t=3\ s\ \; \green{\bigstar}$

So , Orange takes 3 s to reach ground