the earliest drop towers were used for the purpose of making lead shots for muskets. the procedure involved releasing droplets of molten lead from the top of the tower. as the droplets of lead fell through the tower, they formed a spherical droplet. if given enough free fall time, the droplet of molten lead would solidify to form the perfect shot. the tallest of these drop towers in the united states was a 230 - foot tower built in baltimore.
suppose that lead free falls through a drop tower for a distance of 55.9 meters.
a. determine how fast the lead shot is moving after free falling for 55.9 meters.
speed m/s
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b. determine how much time it takes the lead shot to fall this distance of 55.9 meters.
time s
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6. k21q6
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a ball is thrown vertically upwards with a speed of 38.0 m/s.
a. how high does it rise above the point of release?
height m
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b. how much time does it take the ball to reach the peak of its trajectory?
time to reach peak s
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Question

Accepted Answer

### 5. K20Q5
#### a.
# Explanation:
## Step1: Use the kinematic equation $v^{2}=u^{2}+2as$
For free - fall, the initial velocity $u = 0\ m/s$, the acceleration $a=g = 9.8\ m/s^{2}$, and the displacement $s = 55.9\ m$.
Substitute into the equation $v^{2}=0^{2}+2\times9.8\times55.9$
## Step2: Solve for $v$
$$
\begin{align*}
v^{2}&=2\times9.8\times55.9\
v^{2}&=19.6\times55.9\
v^{2}&=1095.64\
v&=\sqrt{1095.64}\
v&\approx33.1\ m/s
\end{align*}
$$
# Answer:
$33.1\ m/s$

#### b.
# Explanation:
## Step1: Use the kinematic equation $s=ut+\frac{1}{2}at^{2}$
Since $u = 0\ m/s$, $s = 55.9\ m$, and $a = 9.8\ m/s^{2}$, the equation becomes $s=\frac{1}{2}at^{2}$
So, $t^{2}=\frac{2s}{a}$
## Step2: Solve for $t$
Substitute $s = 55.9\ m$ and $a = 9.8\ m/s^{2}$ into $t^{2}=\frac{2s}{a}$
$$
\begin{align*}
t^{2}&=\frac{2\times55.9}{9.8}\
t^{2}&=\frac{111.8}{9.8}\
t^{2}& = 11.41\
t&=\sqrt{11.41}\
t&\approx3.38\ s
\end{align*}
$$
# Answer:
$3.38\ s$

### 6. K21Q6
#### a.
# Explanation:
## Step1: Use the kinematic equation $v^{2}=u^{2}+2as$
At the maximum height, $v = 0\ m/s$, the initial velocity $u = 38.0\ m/s$, and the acceleration $a=-g=- 9.8\ m/s^{2}$ (negative because it is against the direction of motion).
Substitute into the equation $0^{2}=38.0^{2}+2\times(-9.8)\times s$
## Step2: Solve for $s$
$$
\begin{align*}
0&=1444-19.6s\
19.6s&=1444\
s&=\frac{1444}{19.6}\
s& = 73.7\ m
\end{align*}
$$
# Answer:
$73.7\ m$

#### b.
# Explanation:
## Step1: Use the kinematic equation $v=u + at$
At the maximum height $v = 0\ m/s$, $u = 38.0\ m/s$, $a=-9.8\ m/s^{2}$
Substitute into $v = u+at$ gives $0 = 38.0-9.8t$
## Step2: Solve for $t$
$$
\begin{align*}
9.8t&=38.0\
t&=\frac{38.0}{9.8}\
t&\approx3.88\ s
\end{align*}
$$
# Answer:
$3.88\ s$

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QUESTION IMAGE

Question

Explanation:

5. K20Q5

a.

Step1: Use the kinematic equation \(v^{2}=u^{2}+2as\)

Step2: Solve for \(v\)

Step1: Use the kinematic equation \(s=ut+\frac{1}{2}at^{2}\)

Step2: Solve for \(t\)

Step1: Use the kinematic equation \(v^{2}=u^{2}+2as\)

Step2: Solve for \(s\)

Answer:

b.