Question

Professor who has a mass of 95kg . Start down a slide that has an angle 40 degree with the horizontal. If the coefficient of kinetic friction between professor's short and the slide is 0.25. What is his acceleration? An airplane is flying 340km/hr at 12 degree east of North. The wind is blowing 40km/hr at 34 degree south of east. What is the palne's actual velocity? Wil-E-Coyole drops a sky bowling ball off a cliff to try to catch the roadrunner. The cliff is 32m high. How long does it take the ball to fall to ground What is it impact velocity ( when the ball hits ground) ? professor, the famous ski jumper leave the ramp with an initial velocity of 34.9 m/s and the angle 35 degree determine total time of flight determine horizontal displacement determine thr peak height. Assume that prof lands at the same height as the top of the ramp and that Li is a projectile

Answer 1

1)

The normal force acting on the box

N = mg cos 40

= (95kg) * 9.8 * cos 40

= 713.19 N

F = mgsin 40 - Ff

where Ff is force of kinetic friction = $\mu$ * N

= 0.25 * 713.19

Ff = 178.29 N

F = mgsin 40 - Ff

ma = mgsin 40 - Ff

(95kg) * a= 95 * 9.8 *sin 40 - 178.29

a = 4.42 m/s²

2)

Velocity component of plane along x and y axes

Vx = (340km/hr) cos 12 = 332.57

Vy = (340km/hr) sin 12 = 70.68

Velocity component of wind along x and y axes

Vx = (40km/hr) cos 34 = 33.16

Vy = (40km/hr) sin 34 = 22.36

Resultant velocity components along x and y axes

V_rx = 332.57-33.16 = 299.41

V_ry = 70.68 - 22.36 = 48.32

Vr = sqrt (V_rx^{2 +} V_ry²)

= 303.28 km/hr

Direction = tan^-1 (V_rx / V_ry)

= tan^-1 (299.41/48.32)

= 80.83

3)

Distance travelled = 0.5 * g * t²

Distance travelled is height = 32m

32 = 0.5 * 9.8 * t²

t= 2.55 sec

Velocity = g * t

= 9.8 * 2.55

= 21.765 m/s

4)

Initial velocity, Vi = 34.9m/s

Angle, $\theta$ = 35

For this question, the vertical velocity must be found to find the total time

Viy = vi sinθ

Viy = 34.9 m/s sin35°

Viy = 20.02 m/s

t= Viy/g

= 20.02 / 9.8

=2.042 sec

b)Horizontal displacement:

Vix = Vi cosθ

= 34.9 m/s cos35°

Vix = 28.58 m/s

dx = Vxt

dx = (28.58m/s)(2.042s)

dx = 58.36m

c) Peak height

With viy, dy can be determined using the following formula (remember that vfy is 0 m/s at maximum height!!!) :

vfy2 = viy2 – 2gdy

(0m/s)² = (20.02 m/s)² – 2(9.81 m/s²)dy

2 * 9.81 dy = 20.02²

dy = 20.42m