A skateboarder starts up a 1.0-m-high, 30∘ ramp at a speed of 7.2 m/s . The skateboard wheels roll without friction. At the top, she leaves the ramp and sails through the air.
How far from the end of the ramp does the skateboarder touch down?
Express your answer with the appropriate units.
Her initial KE-topPE=KE at top
1/2 m 7.2^2 - mg*1m=1/2 m vt^2
vt=sqrt(7.2^2-2*9.8)= 5.68m/s
Now how long does it take to fall 1 meter
hf=hi+vi'*t-4.9t^2
0=1+5.68sin30*t-4.9 t^2
solve that for time t using the quadratic equation (or you calculator).
Now, given the time in air, the horizontal distance is
d=5.68*Cos30*timet
Well, I guess you could say that skateboarder is really getting some "air" time! Let's calculate how far she goes before touching down.
To find the horizontal distance traveled, we can use the equation:
d = v * t
where d is the distance, v is the horizontal component of the velocity, and t is the time.
To find the horizontal component of the velocity, we can use trigonometry:
v = v_initial * cos(theta)
where v_initial is the initial velocity and theta is the angle of the ramp.
Plugging in the given values, we have:
v = 7.2 m/s * cos(30∘)
Now, let's find the time it takes for the skateboarder to reach the peak of the trajectory. We can use the equation:
y = y_initial + v_initial_y * t - (1/2) * g * t^2
where y is the vertical distance, y_initial is the initial vertical position, v_initial_y is the initial vertical component of the velocity, t is the time, and g is the acceleration due to gravity.
Since the skateboarder starts from rest vertically (at the bottom of the ramp), we can simplify the equation to:
y = (1/2) * -g * t^2
where y_initial is 0.
Plugging in the given values and solving for t, we have:
1.0 m = (1/2) * 9.8 m/s^2 * t^2
Solving for t, we find:
t = sqrt(2 * 1.0 m / 9.8 m/s^2) ≈ 0.452 s
Now that we have the time it takes for the skateboarder to reach the peak, we can plug it into the equation for horizontal distance:
d = v * t = (7.2 m/s * cos(30∘)) * 0.452 s
Calculating this, we find:
d ≈ 2.95 m
So, the skateboarder touches down approximately 2.95 meters away from the end of the ramp. Keep on rolling with the calculations!
To find the horizontal distance traveled by the skateboarder after leaving the ramp, we can use the equations of motion.
First, we need to find the initial horizontal velocity of the skateboarder. Since there is no friction, the horizontal velocity remains constant.
Given:
Initial vertical displacement, h = 1.0 m
Ramp angle, θ = 30°
Initial speed, v0 = 7.2 m/s
The horizontal velocity, vx, can be found using the equation:
vx = v0 * cos(θ)
Substituting the given values:
vx = 7.2 m/s * cos(30°)
vx = 7.2 m/s * 0.866
vx = 6.24 m/s
Now, we can find the time of flight, t, using the vertical displacement and acceleration due to gravity. Since the skateboarder leaves the ramp and lands at the same vertical height, the vertical displacement is zero.
Using the equation of motion:
h = v0y * t + (1/2) * g * t^2
Considering the initial vertical velocity is zero (since the skateboarder is only moving horizontally):
0 = (1/2) * g * t^2
Simplifying the equation:
t^2 = 0
t = 0
Since the time is zero, it means the skateboarder takes no time to reach the ground after leaving the ramp.
Therefore, the skateboarder touches down immediately at the end of the ramp.
The horizontal distance traveled by the skateboarder is zero.
To find out how far from the end of the ramp the skateboarder will touch down, we can use the basic principles of projectile motion. We need to find the horizontal distance traveled by the skateboarder during the time it takes for her to reach the ground.
First, let's find the time it takes for the skateboarder to reach the ground. We can use the vertical motion equation:
h = v0y * t + (1/2) * g * t^2
Where:
- h is the vertical distance traveled (1.0 m)
- v0y is the vertical component of the initial velocity (initial vertical speed)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time taken to reach the ground
Since the skateboard wheels roll without friction, there is no vertical force acting on the skateboarder after leaving the ramp. This means v0y remains constant throughout the motion. We can calculate v0y using the initial speed and the angle of the ramp:
v0y = v0 * sin(θ)
where:
- v0 is the initial speed of the skateboarder (7.2 m/s)
- θ is the angle of the ramp (30∘)
Plugging in the values:
v0y = 7.2 m/s * sin(30∘)
v0y = 7.2 m/s * 0.5
v0y = 3.6 m/s
Now, let's find the time it takes for the skateboarder to reach the ground. We can rearrange the first equation to solve for t:
h = v0y * t + (1/2) * g * t^2
0 = (1/2) * g * t^2 + v0y * t - h
This is a quadratic equation in terms of t. We can solve it using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
where:
- a = (1/2) * g
- b = v0y
- c = -h
Plugging in the values:
a = (1/2) * 9.8 m/s^2
b = 3.6 m/s
c = -1.0 m
t = (-3.6 ± sqrt((3.6)^2 - 4 * (1/2) * 9.8 * (-1.0))) / (2 * (1/2) * 9.8)
After solving the quadratic equation, we get two possible values for t: t1 and t2. We can ignore the negative value for time in this case because it doesn't make sense for the skateboarder to have a negative time.
Now that we have the time it takes for the skateboarder to reach the ground, we can find the horizontal distance traveled using the horizontal motion equation:
d = v0x * t
Where:
- d is the horizontal distance traveled
- v0x is the horizontal component of the initial velocity (initial horizontal speed)
- t is the time taken to reach the ground
Since there is no horizontal force acting on the skateboarder, the horizontal component of the initial velocity remains constant throughout the motion. We can calculate v0x using the initial speed and the angle of the ramp:
v0x = v0 * cos(θ)
Plugging in the values:
v0x = 7.2 m/s * cos(30∘)
v0x = 7.2 m/s * √3/2
v0x = 3.6 m/s * √3
Now, we can find the horizontal distance traveled by multiplying v0x by the time t:
d = v0x * t
Plug in the values for v0x and t that we calculated earlier to get the final answer.