Projectile Motion interactive worksheet - Free Printable

Problem 1: Pirate Ship and Cannonballs

#### Given:
- Initial velocity, \( v_i = 82 \, \text{m/s} \)
- Launch angle, \( \theta_i = 27^\circ \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

#### To Find:
1. Time of flight, \( t_R \)
2. Range, \( R \)

#### Solution:

##### Step 1: Calculate the time of flight, \( t_R \)
The formula for the time of flight of a projectile is:
\[
t_R = \frac{2 v_i \sin \theta_i}{g}
\]

Substitute the given values:
\[
t_R = \frac{2 \times 82 \times \sin 27^\circ}{9.8}
\]

First, calculate \( \sin 27^\circ \):
\[
\sin 27^\circ \approx 0.454
\]

Now substitute this value:
\[
t_R = \frac{2 \times 82 \times 0.454}{9.8} = \frac{74.344}{9.8} \approx 7.59 \, \text{s}
\]

So, the time of flight is:
\[
t_R \approx 7.59 \, \text{s}
\]

##### Step 2: Calculate the range, \( R \)
The formula for the range of a projectile is:
\[
R = \frac{v_i^2 \sin 2\theta_i}{g}
\]

Substitute the given values:
\[
R = \frac{(82)^2 \sin (2 \times 27^\circ)}{9.8}
\]

First, calculate \( 2 \times 27^\circ \):
\[
2 \times 27^\circ = 54^\circ
\]

Next, calculate \( \sin 54^\circ \):
\[
\sin 54^\circ \approx 0.809
\]

Now substitute these values:
\[
R = \frac{82^2 \times 0.809}{9.8} = \frac{6724 \times 0.809}{9.8} = \frac{5442.916}{9.8} \approx 555.4 \, \text{m}
\]

So, the range is:
\[
R \approx 555.4 \, \text{m}
\]

Final Answers for Problem 1:

\[
\boxed{7.59 \, \text{s}, 555.4 \, \text{m}}
\]

---

Problem 2: Baseball and Outfielder

#### Given:
- Initial velocity, \( v_i = 40.0 \, \text{m/s} \)
- Launch angle, \( \theta_i = 35.0^\circ \)
- The ball is caught at the same height it was hit.

#### To Find:
1. Distance from the batter when the outfielder catches the ball, \( R \)
2. Time taken to catch the ball, \( t_R \)

#### Solution:

##### Step 1: Calculate the time of flight, \( t_R \)
The formula for the time of flight of a projectile is:
\[
t_R = \frac{2 v_i \sin \theta_i}{g}
\]

Substitute the given values:
\[
t_R = \frac{2 \times 40.0 \times \sin 35.0^\circ}{9.8}
\]

First, calculate \( \sin 35.0^\circ \):
\[
\sin 35.0^\circ \approx 0.574
\]

Now substitute this value:
\[
t_R = \frac{2 \times 40.0 \times 0.574}{9.8} = \frac{45.92}{9.8} \approx 4.69 \, \text{s}
\]

So, the time of flight is:
\[
t_R \approx 4.69 \, \text{s}
\]

##### Step 2: Calculate the range, \( R \)
The formula for the range of a projectile is:
\[
R = \frac{v_i^2 \sin 2\theta_i}{g}
\]

Substitute the given values:
\[
R = \frac{(40.0)^2 \sin (2 \times 35.0^\circ)}{9.8}
\]

First, calculate \( 2 \times 35.0^\circ \):
\[
2 \times 35.0^\circ = 70.0^\circ
\]

Next, calculate \( \sin 70.0^\circ \):
\[
\sin 70.0^\circ \approx 0.9397
\]

Now substitute these values:
\[
R = \frac{40.0^2 \times 0.9397}{9.8} = \frac{1600 \times 0.9397}{9.8} = \frac{1503.52}{9.8} \approx 153.4 \, \text{m}
\]

So, the range is:
\[
R \approx 153.4 \, \text{m}
\]

Final Answers for Problem 2:

\[
\boxed{153.4 \, \text{m}, 4.69 \, \text{s}}
\]