Respuesta :
Given,
The initial velocity with which the calculator was thrown, u=9 m/s
The height of the building, h=23 m
A.
When the calculator reaches the peak height, its velocity will become zero. That is v=0 m/s
And while it is going up the acceleration due to gravity will be acting downward. Thus the acceleration due to gravity will be a negative value.
From the equation of motion,
[tex]v^2-u^2=2gs_{}[/tex]Where s is the peak height reached by the calculator and g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} 0-9^2=2\times-9.8\times s \\ \Rightarrow s=\frac{-9^2}{2\times-9.8} \\ =4.13\text{ m} \end{gathered}[/tex]Thus the total height reached by the calculator from the ground is
[tex]\begin{gathered} H=h+s \\ =23+4.13 \\ =27.13 \end{gathered}[/tex]After reaching the peak height, the calculator starts descending. This descent starts with the initial velocity of v=0 m/s. And the acceleration due to gravity will be in the direction of motion of the calculator. Thus it will be a positive value.
From the equation of motion,
[tex]v^2_0-v^2=2gH[/tex]where v₀ is the velocity of the calculator right before it hits the ground.
On substituting the known values,
[tex]\begin{gathered} v^2_0-0=2\times9.8\times27.13 \\ \Rightarrow v_0=\sqrt[]{2\times9.8\times27.13}^{} \\ =23.06\text{ m/s} \end{gathered}[/tex]Thus the speed of the calculator right before it hits the ground is 23.06 m/s
B.
The acceleration of the calculator is a constant value. It is always equal to the acceleration due to gravity.
Thus the acceleration of the calculator at peak height is 9.8 m/s²
C.
From the equation of motion,
[tex]v=u+gt[/tex]Where t is the time it takes for the calculator to reach the peak height.
On substituting the known values,
[tex]\begin{gathered} 0=9+(-9.8)t \\ \Rightarrow t=\frac{-9}{-9.8} \\ =0.92\text{ s} \end{gathered}[/tex]Thus it takes 0.92 s for the calculator to reach the peak height.
