Question

A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. What is the horizontal velocity component of the ball? Round the answer to the nearest tenth of a m/s

Answer 1

The horizontal component of the velocity of the ball is calculated by multiplying the speed by the cosine of the given angle. 
                        x-component of speed = (31 m/s)(cos 35°) 
                                                               = 25.39 m/s
Thus, the horizontal velocity component of the ball is 25.39 m/s. 

Answer 2

Answer:

Horizontal component of the speed is given as

$v_x = 25.4 m/s$

Explanation:

As we know that if a vector is inclined at some angle with the horizontal then the component of the vector along and perpendicular to horizontal is given as

$A_x = Acos\theta$

$A_y = Asin\theta$

now here we need to find the component of velocity

so we know that

$v = 31 m/s$

$\theta = 35^0$

now we will have

$v_x = 31cos35$

$v_x = 25.4 m/s$

so horizontal component of the speed will be 25.4 m/s