Question

I believe I have the answer for part a, could you help me with part b?

Answer 1

We know that

• The angle is 22 to the vertical.

,

• The vertical component of the velocity is 38 m/s.

First, we have to find the angle to the horizontal. Subtract 22 from 90 degrees.

`\theta=90-22=68`

Then, find the velocity using the vertical component formula.

`\begin{gathered} v_y=v_{}\cdot\sin \theta \\ 38\cdot(m)/(s)=v\cdot\sin 68 \\ v=(38\cdot(m)/(s))/(\sin 68) \\ v\approx41\cdot(m)/(s) \end{gathered}`

Once we have the velocity, we can find the horizontal component of the velocity.

`\begin{gathered} v_x=v\cdot\cos \theta_{} \\ v_x=41\cdot(m)/(s)\cdot\cos 68 \\ v_x\approx15\cdot(m)/(s) \end{gathered}`

(a) Therefore, the horizontal component is 15 m/s.

(b) (i) After one second, the horizontal component will be the same as before because this vector has a constant motion, the gravity doesn't affect the horizontal movement.

(ii) But, in the case of the vertical component, it will be smaller after one second because the gravity slows down this velocity.