Question

A basketball player shoots the ball with an initial velocity of 17.1 ft/sec at an angle of 45.6 degrees with the horizontal.to the nearest tenth,find the initial horizontal component of the vector that represents the velocity of the ball

5.1 ft/sec
12.0 ft/sec
12.2 ft/sec
17.5 ft/sec

i got 5.1 but not sure if correct

Answer 1

The horizontal component is:

17.1cos45.6≈11.96 so

vx=12.0 ft/sec (to nearest tenth ft/s)

Answer 2

Answer: 12.0 ft/sec

Explanation:

From the given question, we recall the following,

A basketball player shoots the ball with an initial velocity = 17.1 ft/sec

With an angle of 45.6 degrees with the horizontal.

Now,

Let us find the initial horizontal component of the vector that represents the velocity of the ball.

cos θ = adjacent/hypotenuse

Cos 45.6 degrees = 17.1 ft/sec

Let find the value of x

x = 17.1 cos 45.6 degrees= 11.96

To the nearest tenth is now = 12.0 ft/sec