Question

a ball is thrown upward with a speed of 29 m/s at an angle of 54° to the horizontal. what is the horizontal component of its velocity 0.50 s later? (air friction is insignificant.) a. 17 m/s b. 20 m/s c. 23 m/s d. 29 m/s v = 23.5 v = 18.6 + (-9.8)(0.5)

Explanation:

Step1: Recall velocity - component formula

The horizontal component of the initial velocity is given by $v_{0x}=v_0\cos\theta$, where $v_0$ is the initial velocity and $\theta$ is the angle of projection.

Step2: Substitute given values

We are given that $v_0 = 29\ m/s$ and $\theta = 54^{\circ}$. So, $v_{0x}=29\cos54^{\circ}$.
We know that $\cos54^{\circ}\approx0.5878$. Then $v_{0x}=29\times0.5878 = 17.0462\approx17\ m/s$.

Answer:

C. 17 m/s