You push a skateboard so that it rolls down the road at a speed of 1.20 m/s. You run after the skateboard at a speed of 3.30 m/s and while still behind the skateboard you jump off the ground at an angle of 28.0° above the horizontal hoping to land on the skateboard. How much distance do you need between you and the skateboard to jump and land on it?
If you do it in relative speed, it is easy..
relative speed=3.3-1.2=2.1 m/s
horizontal distance=2.1cosTheta*time
vertical time in air:
hf=hi+2.1sin28 t-4.8t^2
0=0+t(2.1sin28-4.8t)
t=2.1sun28/4.8 Now put that time in the horiztonal distance, and solve for distance.
To determine the distance you need to jump to land on the skateboard, we can break down the problem into horizontal and vertical components.
Let's start with the horizontal component. Since you are running at a speed of 3.30 m/s, that will be your initial horizontal velocity (Vox). The skateboard is already moving at a speed of 1.20 m/s, so you need to account for that as well. Therefore, the relative velocity between you and the skateboard will be:
Vox = 3.30 m/s - 1.20 m/s
= 2.10 m/s
Now, let's consider the vertical component. The angle at which you jump, 28.0° above the horizontal, gives us the vertical velocity component (Voy). However, we don't know the initial vertical velocity. We can find it using the equation:
Voy = V * sin(θ)
Where V is the initial velocity and θ is the angle.
Since we are jumping off the ground, the initial vertical velocity (Voy) is 0. Therefore, we can solve for the initial velocity (V) and substitute it into the equation for vertical motion:
0 = V * sin(28.0°)
V = 0 / sin(28.0°)
V = 0
This means that you will not have any initial vertical velocity when you jump off the ground. You will only experience the force of gravity pulling you towards the ground.
To land on the skateboard, you need to jump and meet it at the same height. The skateboard has been rolling for some time, and you want to jump and catch up to it. In other words, the vertical displacement of your jump should match the vertical displacement of the skateboard.
The vertical displacement can be found using the equation:
Δy = V * t + (1/2) * g * t^2
Where V is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (9.8 m/s²).
Since we established that the initial vertical velocity (V) is 0, the equation simplifies to:
Δy = (1/2) * g * t^2
The skateboard has been rolling for some time, so we need to find the time it has been in motion (t). The time it takes for you to jump and meet the skateboard can be found using the horizontal distance you need to cover and the relative velocity between you and the skateboard.
Let's denote the horizontal distance as x. The time (t) can be found using the equation:
t = x / Vox
Substituting the known values:
t = x / 2.10 m/s
Now, we can substitute the expression for time (t) into the equation for vertical displacement:
Δy = (1/2) * g * (x / Vox)^2
The vertical displacement (Δy) should be equal to the height of the skateboard above the ground. This height is not given in the question, so we'll assume it's 0.10 m for the sake of calculation.
Therefore, we have:
0.10 m = (1/2) * 9.8 m/s² * (x / 2.10 m/s)^2
Now we can solve for x:
0.10 m = (1/2) * 9.8 m/s² * (x / 2.10 m/s)^2
Simplifying further:
0.10 m = 4.9 m/s² * (x / 2.10 m/s)^2
0.20 m/s² * (x / 2.10 m/s)^2 = 0.10 m
(x / 2.10 m/s)^2 = 0.10 m / (0.20 m/s²)
(x / 2.10 m/s)^2 = 0.50 s²
Taking the square root of both sides:
x / 2.10 m/s = √(0.50 s²)
x = 2.10 m/s * √(0.50 s²)
x ≈ 1.48 m
Therefore, you need to have a distance of approximately 1.48 meters between you and the skateboard in order to jump and land on it.