Question

Jack and Jill ran up the hill at 2.8 m/s . The horizontal component of Jill's velocity vector was 2.1 m/s . What was the angle of the hill?What was the vertical component of Jill's velocity?

Answer 1

Answer:$\theta =41.409 ^{\circ}$

Explanation:

Given

Jack and Jill ran up the hill at 2.8 m/s

Horizontal component of Jill's velocity vector was 2.1 m/s

Let $\theta$is the angle made by Jill's velocity with it's horizontal component

Therefore

$2.8cos\theta =2.1$

$cos\theta =\frac{2.1}{2.8}$

$cos\theta =0.75$

$\theta =41.409 ^{\circ}$

Vertical velocity is given by

$V_y=2.8sin41.11=1.85 m/s$