A skateboarder shoots off a ramp with a velocity of 6.7 m/s, directed at an angl

Question

A skateboarder shoots off a ramp with a velocity of 6.7 m/s, directed at an angle of 57° above the horizontal. The end of the ramp is1.1 m above the ground. Let the xaxis be parallel to the ground, the +y direction bevertically upward, and take as the origin the point on the grounddirectly below the top of the ramp. (a) How high above the ground is the highestpoint that the skateboarder reaches?
Enter anumber.               m.

(b) When the skateboarder reaches the highest point, how far isthis point horizontally from the end of the ramp?
Enter anumber. 2 m (a) How high above the ground is the highestpoint that the skateboarder reaches?
Enter anumber.               m.

(b) When the skateboarder reaches the highest point, how far isthis point horizontally from the end of the ramp?
Enter anumber. 2 m Enter anumber.               m. Enter anumber.

Explanation / Answer

We know that
             v2 - u2 = 2*a*h ==>       0 -(5.619)2 = - 2*9.8*h Therefore the high above the ground is the highest point thatthe skateboarder reaches is                       = 1.1 + 1.611
                      = 2.711 m
The time taken to reach the highest point is
                     t = u / g
                       = 5.619 / 9.8
                       = 0.5733 s
(b)
Initial horizontal velocity u ' = 6.7 *cos57 = 3.649m/s
horizontal acceleration = 0 Therefore the distance traveled in 0.5733 s is
d= u' * t
                    = 3.649 * 0.5733
                    = 2.092 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---