Question

at what initial height must a projectile be launched in order to travel more than 37 m if launched at an angle of 30° with an initial velocity of 20 m/s? 2 m 1 m 0 m 3 m

Explanation:

Step1: Calculate horizontal and vertical components of initial velocity

The initial velocity \(v_0 = 20\ m/s\), and the launch angle \(\theta=30^{\circ}\).
The horizontal component \(v_{0x}=v_0\cos\theta = 20\cos30^{\circ}=20\times\frac{\sqrt{3}}{2} = 10\sqrt{3}\ m/s\).
The vertical component \(v_{0y}=v_0\sin\theta=20\sin30^{\circ}=10\ m/s\).

Step2: Use the kinematic equation for vertical motion \(y = y_0+v_{0y}t-\frac{1}{2}gt^2\) (when \(y = 0\)) and horizontal motion \(x = v_{0x}t\)

From \(x = v_{0x}t\), we have \(t=\frac{x}{v_{0x}}\). Substitute \(t\) into \(y = y_0+v_{0y}t-\frac{1}{2}gt^2\) with \(y = 0\).
\(0=y_0 + v_{0y}\frac{x}{v_{0x}}-\frac{1}{2}g(\frac{x}{v_{0x}})^2\).
We want \(x>37\ m\). Let's substitute \(x = 37\ m\), \(v_{0x}=10\sqrt{3}\ m/s\), \(v_{0y}=10\ m/s\), and \(g = 9.8\ m/s^2\) into the equation \(0=y_0 + v_{0y}\frac{x}{v_{0x}}-\frac{1}{2}g(\frac{x}{v_{0x}})^2\).
First, \(\frac{x}{v_{0x}}=\frac{37}{10\sqrt{3}}\ s\approx\frac{37}{17.32}\ s\approx2.14\ s\).
\(v_{0y}\frac{x}{v_{0x}}=10\times\frac{37}{10\sqrt{3}}=\frac{37}{\sqrt{3}}\approx21.36\).
\(\frac{1}{2}g(\frac{x}{v_{0x}})^2=\frac{1}{2}\times9.8\times(\frac{37}{10\sqrt{3}})^2=\frac{4.9\times1369}{300}\approx22.3\).
Then \(y_0=\frac{1}{2}g(\frac{x}{v_{0x}})^2 - v_{0y}\frac{x}{v_{0x}}\approx22.3 - 21.36\approx1\ m\).

Answer:

1 m