Physics, Motion, show solving steps please

Question

The roof of a two-story house makes an angle of 29° with the horizontal. A ball rolling down the roof rolls off the edge at a speed of 5.1 m/s. The distance to the ground from that point is 5.6 m.

(a) How long is the ball in the air?

? s  

(b) How far from the base of the house does it land?

 ? m  

(c) What is its velocity just before landing? (Let upward be the positive y-direction.)

x-component	m/s
y-component	m/s

Answer 1

Draw a velocity vector diagram with the original velocity off the roof (5.1 m/s) directed downward at a 29° angle. You have to find the horizontal and vertical components as follows: horizontal component = (5.1)cos 29

= 4.46 m/s (v_x); vertical component = (5.1)sin 29 = 2.47 m/s downward, so -2.47 m/s (v_yo)

Next use kinematic h = v_yot + 0.5gt² to find t: -5.6 = (-2.47)t + 0.5(-9.8)t² , rearrange to quadratic form and apply quadratic formula: 4.9t² +2.47t - 5.6 = 0; t = 0.875 s (a)

During this time it is moving horizontally at 4.46 m/s so horizontal distance traveled = (0.875)(4.46) = 3.9 m (b)

(c) x-component is constant = 4.46 m/s

y-component was accelerated downward by gravity so use a kinematic v_yf = v_yo + gt = -2.47 + (-9.8)(0.875)

= -11.0 m/s

Actual landing velocity is the hypotenuse of the triangle formed by the horizontal and final vertical velocity:

=sqrt(4.46² + 11²) = 11.9 m/s generally downward