The drawing shows a skateboarder moving at 7.90 m/s along a horizontal section o

Question

The drawing shows a skateboarder moving at 7.90 m/s along a horizontal section of a track that is slanted upward by theta = 46.0 degree above the horizontal at its end, which is 0.570 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track. " 19118992,The total electric flux from a cubical box 37.0 cm on a side is 3600 N middot m^2/C What charge is enclosed by the box? 19118996,The figure shows a shaft a hand-operated machine. The frictional torque in the journal from bearings at A and B is 15 N-m each. Find the diameter (d) of the shaft (on which the pulley is mounted) using maximum distortion energy criterion. The shaft material is 40C8 steel for which the yield stress in tension is 380 MPa and the factor of safe 19118998,A object of mass 3.00 kg is subject to a force F_x that varies with position as in the figure below. Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J Find the work done by the force on the object as it moves from x = 4.00 m to x = 12.0 m. J Find the work done by the force on the object as it moves from x = 12.0 m to x = 17.0 m. J If the object has a speed of 0.450 m/s at x = 0

Explanation / Answer

Given Data:

Velocity at the bottom of the track, v = 7.90 m/s

Vertical height of the track, h = 0.570 m

Angle of inclination of the track, theta = 46.0 degree

Maximum height above the end of the track, H = ?

Let

Velocity at the end of the track = v'

Potential energy + kinetic energy at the end of the track = Kinetic energy at the bottom of the track

m*g*h + (1/2)*m*v'^2 = (1/2)*m*v^2

g*h + (1/2)*v'^2 = (1/2)*v^2

2*g*h + v'^2 = v^2

v'^2 = v^2 - 2*g*h

v' = sqrt(v^2 - 2*g*h)

= sqrt[ 7.90^2 - (2*9.8*0.570)]

= 7.158 m/s

Vertical component of velocity at the end of the track is

vy = v' * sin(theta)

= 7.158 * sin(46)

= 5.149 m/s

Maximum height above the track is

H = vy^2 / 2*g

= 5.149^2 / ( 2 * 9.8 )

= 1.352 m

The Vertical height above the track, H = 1.352 m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.