Need help very much appreciated....Thanks....A skateboarder shoots off a ramp with a velocity of 6.5 m/s, directed at an angle of 62° above the horizontal. The end of the ramp is 1.4 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp.
(a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
I need help to work a problem like this
To solve this problem, we will use the principles of projectile motion.
(a) First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component (Vx) remains constant throughout the motion and can be found using the equation Vx = V * cos(theta), where V is the initial velocity and theta is the angle of projection. Plugging in the values, we get:
Vx = 6.5 m/s * cos(62°) ≈ 2.75 m/s
The vertical component (Vy) changes due to the effects of gravity. We can find the time of flight (t) using the equation t = (2 * Vy) / g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the skateboarder is at the highest point at the peak of the motion, Vy at this point will be 0. So, using this information, we can find t as follows:
0 = Vy - g * t
0 = Vy - 9.8 * t
Vy = 9.8 * t
Now, we can substitute Vy = 9.8 * t into the equation for the vertical displacement (delta y = Vy * t + (1/2) * g * t^2) to find the highest point above the ground. Substituting the values, we get:
1.4 m = (9.8 * t * t) / 2
2.8 = 9.8 * t^2
t^2 = 2.8 / 9.8
t ≈ 0.53 s
Now, we can substitute this value of t into Vy = 9.8 * t to find the vertical component of the initial velocity:
Vy = 9.8 * 0.53 ≈ 5.19 m/s
Finally, to find the highest point above the ground, we use the equation:
delta y = Vy * t + (1/2) * g * t^2
delta y = 5.19 * 0.53 + (1/2) * 9.8 * 0.53^2
delta y ≈ 1.38 m
Therefore, the highest point above the ground that the skateboarder reaches is approximately 1.38 meters.
(b) To find the horizontal distance, we use the equation:
delta x = Vx * t
delta x = 2.75 * 0.53
delta x ≈ 1.46 m
Therefore, when the skateboarder reaches the highest point, the horizontal distance from the end of the ramp is approximately 1.46 meters.