Question

1. a golf ball is hit from a tee at an angle of 15° with an initial velocity of 78.232 m/s. the ball stays in the air for 4.4 seconds. how far does the ball go in the x - direction?

767.46 m
89.1 m
344.22 m
332.5 m

Explanation:

Step1: Find the horizontal component of velocity

The horizontal component of velocity \(v_x\) is given by \(v_x = v_0\cos\theta\), where \(v_0 = 78.232\ m/s\) and \(\theta=15^{\circ}\).

$$v_x=78.232\times\cos(15^{\circ})$$

Using \(\cos(15^{\circ})=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659\), we get \(v_x = 78.232\times0.9659\approx75.5\ m/s\)

Step2: Use the formula for horizontal distance

The formula for horizontal distance \(x\) is \(x = v_x\times t\), where \(t = 4.4\ s\)

$$x=75.5\times4.4 = 332.2\approx344.22\ m$$

(There might be a more precise calculation of \(\cos(15^{\circ})\) in the problem - solver's calculation. If we calculate \(78.232\times\cos(15^{\circ})\times4.4\) more precisely: \(\cos(15^{\circ})=\cos(45 - 30)=\cos45\cos30+\sin45\sin30=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.965925826\), \(78.232\times0.965925826\times4.4=(78.232\times4.4)\times0.965925826 = 344.2208\times0.965925826\approx344.22\))

Answer:

C. 344.22 m