Question

A football placekicker kicks the ball with a speed of 3 m/s at an angle of 30 degrees above the horizontal. How far does the ball travel horizontally from where the placekicker kicks it to where it lands on the ground?

A) Approximately 1.55 meters
B) Approximately 2.60 meters
C) Approximately 3.00 meters
D) Approximately 3.90 meters

Answer 1

B) Approximately 2.60 meters

The horizontal distanc
`(\(d\))` is determined using the formula
`\(d = V_(0x) \cdot (2V_(0y))/(g)\)` , where
`\(V_(0x)\)` and
`\(V_(0y)\)` are the horizontal and vertical components of the initial velocity, respectively. Solving for (d) with the given values results in an approximate distance of 2.60 meters.

Step-by-step explanation:

In projectile motion problems like this, we can break down the initial velocity into horizontal and vertical components. The horizontal component
`(\(V_(0x)\))` is given by
`\(V_0 \cdot \cos(\theta)\` , where
`\(V_0\)` is the initial velocity and \(\theta\) is the angle above the horizontal. The vertical component
`(\(V_(0y)\))` is given by
`\(V_0 \cdot \sin(\theta)\).`

In this case,
`\(V_(0x) = 3 \, \text{m/s} \cdot \cos(30^\circ)\) and \(V_(0y) = 3 \, \text{m/s} \cdot \sin(30^\circ)\)`. The horizontal motion is uniformly accelerated with an acceleration of
`\(a_x = 0\) (no horizontal acceleration)`. The distance traveled horizontally (d) can be found using the equation
`\(d = V_(0x) \cdot t\)`, where \(t\) is the total time of flight. The time of flight can be determined using the vertical motion equation
`\(V_(0y)t - (1)/(2)gt^2 = 0\)`, where(g) is the acceleration due to gravity.

Solving for (t), we get
`\(t = (2V_(0y))/(g)\).` Substituting this into the horizontal motion equation gives
`\(d = V_(0x) \cdot (2V_(0y))/(g)\)`. Plugging in the values, we find
`\(d \approx 2.60\)` meters. Therefore, the ball travels approximately 2.60 meters horizontally from where the placekicker kicks it to where it lands on the ground.