Question

A ball is thrown vertically upwards with a velocity v and an initial kinetic energy Ek. When half way to the top of its flight, it has a velocity and kinetic energy respectively of

A. (v/2) (Ek/2)
B. (v/√2) (Ek/2)

Answer 1

Answer:

The answer is B. (v/√2) (Ek/2).

Explanation:

We can find the velocity and kinetic energy of the ball when it is half way to the top of its flight by using the Mechanical Energy formula:

$\boxed{Mechanical\ Energy\ (ME)=Potential\ Energy\ (PE)+Kinetic\ Energy\ (KE)}$

while,

• $PE=mass\ (m)\times gravity\ acceleration\ (g)\times height\ (h)$
• $KE=\frac{1}{2} \times mass\ (m)\times velocity^2\ (v^2)$

Therefore, we can also write:

$\boxed{ME=mgh+\frac{1}{2} mv^2}$

According to the principle of conservation of mechanical energy, the values of the mechanical energy are constant at all points (provided there is no external energy applied to the system). Therefore the ME at the initial point, the top point or the half way (mid point) will always be the same.

$\boxed{ME_1=ME_2=ME_3=...}$

Let:

• $v_o$ = initial velocity
• $h_o$ = initial height, which we take it as 0
• $v_1$ = velocity at top point, which is equals to 0
• $h_1$ = height of top point
• $v_2$ = velocity of mid point

$ME_0=ME_1$

$mgh_o+\frac{1}{2} mv_o^2=mgh_1+\frac{1}{2} mv_1^2$

($h_o$ = 0 and $v_1$ = 0)

$0+\frac{1}{2} mv_o^2=mgh_1+0$

$\displaystyle h_1=\frac{v_o^2}{2g}$

Since the height of half way is half of the top point, hence:

$h_2=\frac{1}{2} h_1$

$\displaystyle h_2=\frac{1}{2} \times\frac{v_o^2}{2g}$

$\displaystyle h_2=\frac{v_o^2}{4g}$

$ME_0=ME_2$

$mgh_o+\frac{1}{2} mv_o^2=mgh_2+\frac{1}{2} mv_2^2$

$0+\frac{1}{2} v_o^2=gh_2+\frac{1}{2} v_2^2$

$\displaystyle \frac{1}{2} v_o^2=g\times\frac{v_o^2}{4g}+\frac{1}{2} v_2^2$

$2v_2^2=2v_o^2- v_o^2$

$\displaystyle v_2^2=\frac{v_o^2}{2}$

$\boxed{\displaystyle v_2=\frac{v_o}{\sqrt{2} }}$

$KE_0:KE_2=\frac{1}{2} mv_o^2:\frac{1}{2} mv_2^2$

$Ek:KE_2=v_o^2:v_2^2$

$\displaystyle Ek:KE_2=v_o^2:\left(\frac{v_o}{\sqrt{2} \right)} ^2$

$\displaystyle Ek:KE_2=v_o^2:\frac{v_o^2}{2}$

$\displaystyle Ek:KE_2=1:\frac{1}{2}$

$\boxed{\displaystyle KE_2=\frac{Ek}{2}}$