A skateboarder shoots off a ramp with avelocity of 5.5 m/s, directed at an angle
ID: 1681233 • Letter: A
Question
A skateboarder shoots off a ramp with avelocity of 5.5 m/s, directed at an angle of 61° above thehorizontal. The end of the ramp is 1.1 m above the ground. Letthexaxis be parallel tothe ground, the +ydirection be vertically upward, and take as the origin thepoint on the ground directly below the top of theramp. (a)How high above the ground is thehighest point that the skateboarderreaches? (b)When the skateboarder reaches thehighest point, how far is this point horizontally from the end oftheramp? A skateboarder shoots off a ramp with avelocity of 5.5 m/s, directed at an angle of 61° above thehorizontal. The end of the ramp is 1.1 m above the ground. Letthexaxis be parallel tothe ground, the +ydirection be vertically upward, and take as the origin thepoint on the ground directly below the top of theramp. (a)How high above the ground is thehighest point that the skateboarderreaches? (b)When the skateboarder reaches thehighest point, how far is this point horizontally from the end oftheramp? (a)How high above the ground is thehighest point that the skateboarderreaches? (b)When the skateboarder reaches thehighest point, how far is this point horizontally from the end oftheramp? (b)When the skateboarder reaches thehighest point, how far is this point horizontally from the end oftheramp?Explanation / Answer
We know thatv2 - u2 = 2*a*h ==> 0 -(4.8104)2 = - 2*9.8*h Therefore the high above the ground is the highest point thatthe skateboarder reaches is = 1.1 + 1.1806
= 2.2806 m
The time taken to reach the highest point is
t = u / g
= 4.8104 / 9.8
= 0.49085 s
(b)
Initial horizontal velocity u ' = 5.5 *cos61 = 2.666m/s
horizontal acceleration = 0 Therefore the distance traveled in 0.49085 s is
d= u' * t
= 2.666 * 0.49085
= 1.3088 m