Question

a student launches a cannonball at an angle. neglect drag.
how long did it take for the cannonball to reach the ground?
3.8 s
what was the vertical component of the launch speed of the cannonball?
unit
what was the peak height that the cannonball reached?
unit
what was the cannonballs range?
unit
what was the horizontal component of the launch speed of the cannonball?
unit
what was the overall launch speed of the cannonball?
unit
what was the launch angle of the cannonball?
unit
(there is a chart on the right showing a horizontal axis from 0m to 90m with a blue dot at 0m and a green dot at 70m, and some control buttons above the chart)

Explanation:

1. Vertical component of launch speed ($v_y$)

We use the vertical motion equation for time to reach the ground (total time $T = 3.8\,\text{s}$). The time to reach peak is $t = \frac{T}{2} = 1.9\,\text{s}$. At peak, vertical velocity is $0$, so $v_y = g \cdot t$ (using $g = 9.8\,\text{m/s}^2$).

Step1: Calculate time to peak

$t = \frac{3.8}{2} = 1.9\,\text{s}$

Step2: Find $v_y$

$v_y = 9.8 \cdot 1.9 = 18.62\,\text{m/s}$

2. Peak height ($h$)

Use $h = v_y \cdot t - \frac{1}{2} g t^2$. Substitute $v_y = 18.62\,\text{m/s}$, $t = 1.9\,\text{s}$, $g = 9.8\,\text{m/s}^2$.

Step1: Substitute values

$h = 18.62(1.9) - \frac{1}{2}(9.8)(1.9)^2$

Step2: Calculate

$h = 35.378 - 17.689 = 17.689\,\text{m} \approx 17.7\,\text{m}$

3. Cannonball’s range ($R$)

From the graph, the range is the horizontal distance (from $0\,\text{m}$ to $70\,\text{m}$), so $R = 70\,\text{m}$.

4. Horizontal component of launch speed ($v_x$)

Horizontal motion is constant velocity, so $v_x = \frac{R}{T}$. Substitute $R = 70\,\text{m}$, $T = 3.8\,\text{s}$.

Step1: Apply formula

$v_x = \frac{70}{3.8} \approx 18.42\,\text{m/s}$

5. Overall launch speed ($v$)

Use Pythagorean theorem: $v = \sqrt{v_x^2 + v_y^2}$. Substitute $v_x \approx 18.42\,\text{m/s}$, $v_y \approx 18.62\,\text{m/s}$.

Step1: Calculate squares

$v_x^2 \approx 339.3$, $v_y^2 \approx 346.7$

Step2: Sum and square root

$v = \sqrt{339.3 + 346.7} = \sqrt{686} \approx 26.2\,\text{m/s}$

6. Launch angle ($\theta$)

Use $\tan\theta = \frac{v_y}{v_x}$. Substitute $v_y \approx 18.62\,\text{m/s}$, $v_x \approx 18.42\,\text{m/s}$.

Step1: Calculate tangent

$\tan\theta = \frac{18.62}{18.42} \approx 1.011$

Step2: Find angle

$\theta = \arctan(1.011) \approx 45.3^\circ$

Final Answers (with units):

• Vertical component: $\boldsymbol{18.6\,\text{m/s}}$ (or $18.6\,\text{m/s}$)
• Peak height: $\boldsymbol{17.7\,\text{m}}$
• Range: $\boldsymbol{70\,\text{m}}$
• Horizontal component: $\boldsymbol{18.4\,\text{m/s}}$
• Overall speed: $\boldsymbol{26.2\,\text{m/s}}$
• Launch angle: $\boldsymbol{45.3^\circ}$ (or $\boldsymbol{45^\circ}$ approx)

(Note: Units are assumed as $\text{m/s}$ for speed, $\text{m}$ for height, $\text{m}$ for range, and degrees for angle, consistent with projectile motion conventions.)

Answer:

1. Vertical component of launch speed ($v_y$)

We use the vertical motion equation for time to reach the ground (total time $T = 3.8\,\text{s}$). The time to reach peak is $t = \frac{T}{2} = 1.9\,\text{s}$. At peak, vertical velocity is $0$, so $v_y = g \cdot t$ (using $g = 9.8\,\text{m/s}^2$).

Step1: Calculate time to peak

$t = \frac{3.8}{2} = 1.9\,\text{s}$

Step2: Find $v_y$

$v_y = 9.8 \cdot 1.9 = 18.62\,\text{m/s}$

2. Peak height ($h$)

Use $h = v_y \cdot t - \frac{1}{2} g t^2$. Substitute $v_y = 18.62\,\text{m/s}$, $t = 1.9\,\text{s}$, $g = 9.8\,\text{m/s}^2$.

Step1: Substitute values

$h = 18.62(1.9) - \frac{1}{2}(9.8)(1.9)^2$

Step2: Calculate

$h = 35.378 - 17.689 = 17.689\,\text{m} \approx 17.7\,\text{m}$

3. Cannonball’s range ($R$)

From the graph, the range is the horizontal distance (from $0\,\text{m}$ to $70\,\text{m}$), so $R = 70\,\text{m}$.

4. Horizontal component of launch speed ($v_x$)

Horizontal motion is constant velocity, so $v_x = \frac{R}{T}$. Substitute $R = 70\,\text{m}$, $T = 3.8\,\text{s}$.

Step1: Apply formula

$v_x = \frac{70}{3.8} \approx 18.42\,\text{m/s}$

5. Overall launch speed ($v$)

Use Pythagorean theorem: $v = \sqrt{v_x^2 + v_y^2}$. Substitute $v_x \approx 18.42\,\text{m/s}$, $v_y \approx 18.62\,\text{m/s}$.

Step1: Calculate squares

$v_x^2 \approx 339.3$, $v_y^2 \approx 346.7$

Step2: Sum and square root

$v = \sqrt{339.3 + 346.7} = \sqrt{686} \approx 26.2\,\text{m/s}$

6. Launch angle ($\theta$)

Use $\tan\theta = \frac{v_y}{v_x}$. Substitute $v_y \approx 18.62\,\text{m/s}$, $v_x \approx 18.42\,\text{m/s}$.

Step1: Calculate tangent

$\tan\theta = \frac{18.62}{18.42} \approx 1.011$

Step2: Find angle

$\theta = \arctan(1.011) \approx 45.3^\circ$

Final Answers (with units):

• Vertical component: $\boldsymbol{18.6\,\text{m/s}}$ (or $18.6\,\text{m/s}$)
• Peak height: $\boldsymbol{17.7\,\text{m}}$
• Range: $\boldsymbol{70\,\text{m}}$
• Horizontal component: $\boldsymbol{18.4\,\text{m/s}}$
• Overall speed: $\boldsymbol{26.2\,\text{m/s}}$
• Launch angle: $\boldsymbol{45.3^\circ}$ (or $\boldsymbol{45^\circ}$ approx)

(Note: Units are assumed as $\text{m/s}$ for speed, $\text{m}$ for height, $\text{m}$ for range, and degrees for angle, consistent with projectile motion conventions.)