Question

a student launches a cannonball horizontally from a tower. neglect drag.
how long did it take for the cannonball to reach the ground?
2.8 s
what was the launch height of the cannonball?
m
what was the cannonball’s range?
m
what was the launch speed of the cannonball?
m/s
(chart: a vertical bar at 0m, a dot at 70m - 80m range, scale 0m - 90m)

Explanation:

1. Time to reach the ground

The cannonball is launched horizontally, so the vertical motion is free - fall. The time to fall from a height \(h\) is given by the equation for free - fall: \(h=\frac{1}{2}gt^{2}\), and we can also use the fact that if we know the time of flight, for vertical motion with initial vertical velocity \(v_{0y} = 0\) (since it's launched horizontally), the time \(t\) can be found from the vertical displacement. However, from the problem, we are given that the time to reach the ground is \(t = 2.8\space s\) (as seen in the green box for the first question). But if we want to derive it, for vertical motion:
We know that in vertical direction, the displacement \(y=h=\frac{1}{2}gt^{2}\), and if we solve for \(t\), \(t=\sqrt{\frac{2h}{g}}\). But since the problem might have given us the time as \(2.8\space s\) (maybe from a simulation or previous calculation), the time to reach the ground \(t = 2.8\space s\).

2. Launch height of the cannonball

For vertical motion (free - fall), the height \(h\) (launch height) is given by the formula \(h=\frac{1}{2}gt^{2}\), where \(g = 9.8\space m/s^{2}\) and \(t = 2.8\space s\).

Step 1: Identify the formula for vertical displacement in free - fall

The formula for the vertical displacement \(y\) (which is equal to the launch height \(h\) when the cannonball reaches the ground) for an object with initial vertical velocity \(v_{0y}=0\) is \(h=\frac{1}{2}gt^{2}\).

Step 2: Substitute the values of \(g\) and \(t\)

Substitute \(g = 9.8\space m/s^{2}\) and \(t = 2.8\space s\) into the formula:
\(h=\frac{1}{2}\times9.8\times(2.8)^{2}\)
First, calculate \((2.8)^{2}=7.84\)
Then, \(\frac{1}{2}\times9.8\times7.84 = 4.9\times7.84\)
\(4.9\times7.84 = 38.416\space m\approx38.4\space m\)

3. Cannonball's range

The range \(R\) of a horizontally launched projectile is given by the formula \(R = v_{0x}\times t\), where \(v_{0x}\) is the initial horizontal velocity (launch speed) and \(t\) is the time of flight. From the graph, the range (horizontal distance traveled) is up to \(70\space m\) (looking at the position of the cannonball on the x - axis). But if we calculate it using the formula, and if we assume that we will find \(v_{0x}\) later, and we know \(t = 2.8\space s\), and from the graph the range \(R\approx70\space m\) (since the cannonball lands at \(x = 70\space m\) approximately).

4. Launch speed of the cannonball

The launch speed \(v_{0}\) is the horizontal speed \(v_{0x}\) (since it's launched horizontally, \(v_{0y} = 0\)). The range \(R=v_{0x}\times t\), so \(v_{0x}=\frac{R}{t}\). If \(R = 70\space m\) and \(t = 2.8\space s\)

Step 1: Identify the formula for horizontal motion

In horizontal motion (neglecting air resistance), the horizontal velocity \(v_{0x}\) is constant, and the range \(R=v_{0x}\times t\), so \(v_{0x}=\frac{R}{t}\)

Step 2: Substitute the values of \(R\) and \(t\)

Substitute \(R = 70\space m\) and \(t = 2.8\space s\) into the formula: \(v_{0x}=\frac{70}{2.8}=25\space m/s\)

Final Answers:

• Time to reach the ground: \(\boldsymbol{2.8\space s}\)
• Launch height: \(\boldsymbol{38.4\space m}\) (using \(h=\frac{1}{2}\times9.8\times(2.8)^{2}\))
• Range: \(\boldsymbol{70\space m}\) (from the graph)
• Launch speed: \(\boldsymbol{25\space m/s}\) (using \(v_{0x}=\frac{70}{2.8}\))

Answer:

1. Time to reach the ground

The cannonball is launched horizontally, so the vertical motion is free - fall. The time to fall from a height \(h\) is given by the equation for free - fall: \(h=\frac{1}{2}gt^{2}\), and we can also use the fact that if we know the time of flight, for vertical motion with initial vertical velocity \(v_{0y} = 0\) (since it's launched horizontally), the time \(t\) can be found from the vertical displacement. However, from the problem, we are given that the time to reach the ground is \(t = 2.8\space s\) (as seen in the green box for the first question). But if we want to derive it, for vertical motion:
We know that in vertical direction, the displacement \(y=h=\frac{1}{2}gt^{2}\), and if we solve for \(t\), \(t=\sqrt{\frac{2h}{g}}\). But since the problem might have given us the time as \(2.8\space s\) (maybe from a simulation or previous calculation), the time to reach the ground \(t = 2.8\space s\).

2. Launch height of the cannonball

For vertical motion (free - fall), the height \(h\) (launch height) is given by the formula \(h=\frac{1}{2}gt^{2}\), where \(g = 9.8\space m/s^{2}\) and \(t = 2.8\space s\).

Step 1: Identify the formula for vertical displacement in free - fall

The formula for the vertical displacement \(y\) (which is equal to the launch height \(h\) when the cannonball reaches the ground) for an object with initial vertical velocity \(v_{0y}=0\) is \(h=\frac{1}{2}gt^{2}\).

Step 2: Substitute the values of \(g\) and \(t\)

Substitute \(g = 9.8\space m/s^{2}\) and \(t = 2.8\space s\) into the formula:
\(h=\frac{1}{2}\times9.8\times(2.8)^{2}\)
First, calculate \((2.8)^{2}=7.84\)
Then, \(\frac{1}{2}\times9.8\times7.84 = 4.9\times7.84\)
\(4.9\times7.84 = 38.416\space m\approx38.4\space m\)

3. Cannonball's range

The range \(R\) of a horizontally launched projectile is given by the formula \(R = v_{0x}\times t\), where \(v_{0x}\) is the initial horizontal velocity (launch speed) and \(t\) is the time of flight. From the graph, the range (horizontal distance traveled) is up to \(70\space m\) (looking at the position of the cannonball on the x - axis). But if we calculate it using the formula, and if we assume that we will find \(v_{0x}\) later, and we know \(t = 2.8\space s\), and from the graph the range \(R\approx70\space m\) (since the cannonball lands at \(x = 70\space m\) approximately).

4. Launch speed of the cannonball

The launch speed \(v_{0}\) is the horizontal speed \(v_{0x}\) (since it's launched horizontally, \(v_{0y} = 0\)). The range \(R=v_{0x}\times t\), so \(v_{0x}=\frac{R}{t}\). If \(R = 70\space m\) and \(t = 2.8\space s\)

Step 1: Identify the formula for horizontal motion

In horizontal motion (neglecting air resistance), the horizontal velocity \(v_{0x}\) is constant, and the range \(R=v_{0x}\times t\), so \(v_{0x}=\frac{R}{t}\)

Step 2: Substitute the values of \(R\) and \(t\)

Substitute \(R = 70\space m\) and \(t = 2.8\space s\) into the formula: \(v_{0x}=\frac{70}{2.8}=25\space m/s\)

Final Answers:

• Time to reach the ground: \(\boldsymbol{2.8\space s}\)
• Launch height: \(\boldsymbol{38.4\space m}\) (using \(h=\frac{1}{2}\times9.8\times(2.8)^{2}\))
• Range: \(\boldsymbol{70\space m}\) (from the graph)
• Launch speed: \(\boldsymbol{25\space m/s}\) (using \(v_{0x}=\frac{70}{2.8}\))