A ball is shot from a compressed-air gun at twice its terminal speed.

a. What is the ball's initial acceleration, as a multiple of $g$, if it is shot straight up?

b. What is the ball's initial acceleration, as a multiple of $g$, if it is shot straight down?

Balls speed is twice its initial speed: $2v_{t.}$

Ball is shot up.

$$\begin{gathered} F=ma \\ {F}_d=mg\frac{v^2}{{v^2}_{t_{}}} \end{gathered}$$

The formula for the drag force is given as:

$F_d=mg\frac{v^2}{v_t^2}.$

Where, $V$is ball's speed

And$v_t$is terminal speed.

When the ball is shot with twice of its terminal speed, $2$forces acting simultaneously on it.

1.$F_d=mg\frac{v^2}{v_t^2}$(drag force opposite in the direction of motion)

2.$F_g=mg$(Earth's gravitational pull always in downward direction)

So, the net force on the ball in the vertical direction can be given as

$\begin{array}{rl}& \sum F_y=F_d-F_g\\ & \sum F_y=mg\frac{v^2}{v_t^2}-mg\end{array}$

Speed of the ball = twice the terminal speed of the ball$v=2v_t$

So,

$\begin{array}{rl}& \sum F_y=mg\frac{{\left(2v_t\right)}^2}{v_t^2}-mg\\ & \sum F_y=4mg-mg\\ & ma=3mg\\ & a=3g\end{array}$

So, the Ball's initial acceleration, if it is shot straight up is $3g$.

Balls speed is twice its initial speed:$2v_{t.}$

Ball is shot up.

$$\begin{gathered} F=ma \\ {F}_d=mg\frac{v^2}{{v^2}_{t_{}}} \end{gathered}$$

The formula for the drag force is given as

$F_d=mg\frac{v^2}{v_{\mathrm{r}}^2}\text{.}$

Where, $V$is ball's speed

And, $v_t$is terminal speed.

When the ball is shot with twice of its terminal speed, 2 forces acting simultaneously on it.

1. $F_d=mg\frac{v^2}{v_t^2}$(drag force opposite in the direction of motion)

2. $F_g=mg$(Earth's gravitational pull always in downward direction)

So, the net force on the ball in vertical direction can be given as:

$\begin{array}{rl}& \sum F_y=-F_d-F_g\\ & \sum F_y=-mg\frac{v^2}{v_t^2}-mg.\end{array}$

Speed of the ball $=$twice the terminal speed of the ball $v=2v_t.$

So,

$\begin{array}{rl}& \sum F_y=-mg\frac{{\left(2v_t\right)}^2}{v_x^2}-mg\\ & \sum F_y=-4mg-mg\\ & ma=-5mg\\ & a=-5g\end{array}$

So, the Ball's initial acceleration, if it is shot straight up is $-5g$.

Part (a): The Ball's initial acceleration, if it is shot straight up is $3g.$

Part (b): The Ball's initial acceleration, if it is shot straight up is$-5g.$