You push a skateboard so that it rolls down the road at a speed of 0.600 m/s. You run after the skateboard at a speed of 3.50 m/s and while still behind the skateboard you jump off the ground at an angle of 22.0° 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?
how far can you jump with that speed and angle? The range is
R = v^2/g sin2θ
Now that you know the distance, figure your time in the air, with a constant speed of v cosθ.
Now, how far does the board go in that time?
You can't be farther away than the difference in distances.
To determine the distance you need to jump in order to land on the skateboard, we can break down the motion into horizontal and vertical components.
First, let's calculate how far the skateboard would have traveled by the time you jump. We'll consider the time it takes for you to jump as 't'.
The horizontal distance traveled by the skateboard would be given by:
Distance = Speed × Time
Distance = 0.600 m/s × t
Now, let's calculate your distance from the skateboard when you jump. This can be achieved by finding the distance you've covered in the horizontal direction while running at a speed of 3.50 m/s.
Distance = Speed × Time
Distance = 3.50 m/s × t
Since we want to land on the skateboard, our target distance would be the difference between these two distances.
Target Distance = Distance from running - Distance traveled by the skateboard
Target Distance = (3.50 m/s × t) - (0.600 m/s × t)
Target Distance = (3.50 - 0.600) m/s × t
Target Distance = 2.90 m/s × t
Now, let's find the time 't'. We'll use the vertical component of motion to calculate it.
When you jump, the vertical distance covered by you can be given by:
Vertical Distance = Initial Vertical Velocity × Time + 0.5 × Acceleration × Time^2
Since you jump off the ground, your initial vertical velocity would be 0 m/s, and the acceleration due to gravity (neglecting air resistance) is -9.8 m/s^2 (negative since it opposes your upward motion).
Considering the vertical component, the equation becomes:
0 = 0 + 0.5 × (-9.8 m/s^2) × t^2
0 = -4.9 m/s^2 × t^2
Simplifying the equation, we get:
t^2 = 0
This shows that the time 't' will be zero, indicating that you will never land on the skateboard if you jump at an angle of 22.0° above the horizontal.
Therefore, you will need to be directly above the skateboard to successfully land on it.