Question

how far will a projectile travel if launched at a 37° angle with an initial height of 0 m and an initial velocity of 24 m/s? 56.44 m 35.89 m 23.90 m 68.42 m

Explanation:

Step1: Find the time of flight

The vertical component of the initial velocity is \(v_{0y}=v_0\sin\theta\), where \(v_0 = 24\ m/s\) and \(\theta = 37^{\circ}\). So \(v_{0y}=24\sin37^{\circ}\approx24\times0.6 = 14.4\ m/s\).
Using the equation \(y = v_{0y}t-\frac{1}{2}gt^2\) (since \(y = 0\) at launch and landing), \(0=14.4t - 4.9t^2\). Factoring out \(t\): \(t(14.4 - 4.9t)=0\). One solution is \(t = 0\) (launch time), and the other is \(t=\frac{14.4}{4.9}\approx2.94\ s\).

Step2: Find the horizontal distance

The horizontal component of the initial velocity is \(v_{0x}=v_0\cos\theta\), \(v_{0x}=24\cos37^{\circ}\approx24\times0.8 = 19.2\ m/s\).
Using the equation \(x = v_{0x}t\), substituting \(v_{0x}=19.2\ m/s\) and \(t = 2.94\ s\), \(x=19.2\times2.94 = 56.44\ m\).

Answer:

56.44 m