Respuesta :
Given that,
Time = 0.5 s
Acceleration = 10 m/s²
(I). We need to calculate the speed of apple
Using equation of motion
[tex]v=u+at[/tex]
Where, v = speed
u = initial speed
a = acceleration
t = time
Put the value into the formula
[tex]v=0+10\times0.5[/tex]
[tex]v=5\ m/s[/tex]
(III). We need to calculate the height of the branch of the tree from the ground
Using equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]s=0+\dfrac{1}{2}\times10\times(0.5)^2[/tex]
[tex]s=1.25\ m[/tex]
(II). We need to calculate the average velocity during 0.5 sec
Using formula of average velocity
[tex]v_{avg}=\dfrac{\Delta x}{\Delta t}[/tex]
[tex]v_{avg}=\dfrac{x_{f}-x_{i}}{t_{f}-t_{0}}[/tex]
Where, [tex]x_{f}[/tex]= final position
[tex]x_{i}[/tex] = initial position
Put the value into the formula
[tex]v_{avg}=\dfrac{1.25+0}{0.5}[/tex]
[tex]v_{avg}=2.5\ m/s[/tex]
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.
