Projectile race track physics help

In summary: Once you have vi, you can use trigonometry to find the horizontal component of velocity. Then, you can use the formula v=vi + at to find the horizontal distance travelled. Finally, you can use the equation y=yi + vi t + 0.5 a t^2 to find the maximum height. In summary, the rider in the racing event jumped off a slope at 30° with a height of 1m and remained in mid-air for 1 second. Using the equations v=vi + at and y=yi + vi t + 0.5 a t^2, the speed at which he was travelling off the slope can be determined, as well as the horizontal distance he travelled before striking
  • #1
Sam Fred
15
0

Homework Statement


The Track for this racing event was designed so that riders jump off the slope at 30°, from a
height of 1m. During a race it was observed that the rider shown in Fig. 2 remained in mid air for
1s. Determine,
1) the speed at which he was traveling off the slope, ( 2points)
2) the horizontal distance he travels before striking the ground, (2points)
3) the maximum height above the ground he attains (3points)

Homework Equations


v=vi + at
y=yi + vi t + 0.5 a t^2

The Attempt at a Solution


in the attachment .
 

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  • #2


hey man, welcome to physicsforums! When they say he remained in mid-air for 1 second, I think this means that he was airborne (not touching ground) for 1 second. Not that he was half-way through his jump at 1 second.
 
  • #3


aha ... thanks man ... but how would i know the speed off he slope ?
Do i repeat the same steps with substituting t = 0.5 sec
 
  • #4


No. That would work if his jump started and finished at the same height. But in this problem, it doesn't, because he starts the jump at 1m height, and finishes the jump at zero metres. So at t=0.5 sec, he is not at the highest part of the jump (he has already gone past that point). So the time when he has zero velocity will be some value, which is greater than 0.5 seconds.

You can use this equation you wrote: y=yi + vi t + 0.5 a t^2 (also, remember that vi needs to be the vertical component of initial velocity). You know the total change in height and you know the change in time, so you can simply solve for vi
 
  • #5


I would approach this problem by first analyzing the given information and identifying the relevant equations that can be used to solve for the unknown variables. From the given information, we know that the rider jumps off the slope at an angle of 30° from a height of 1m, and remains in mid-air for 1s. We can use the equations for projectile motion to solve for the unknowns.

1) To determine the speed at which the rider was traveling off the slope, we can use the equation v=vi + at, where v is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. Since the rider remains in mid-air for 1s, we can substitute t=1s. The acceleration in the y-direction is due to gravity and is equal to -9.8 m/s^2. The initial velocity in the y-direction is 0 m/s, since the rider starts from rest. Therefore, we can rearrange the equation to solve for the final velocity: v = 0 + (-9.8)(1) = -9.8 m/s. However, since the rider is moving in a parabolic path, we also need to consider the horizontal velocity component. The angle of 30° tells us that the horizontal and vertical velocities are related by a trigonometric function, in this case, sine. So, using the trigonometric relationship, we can calculate the horizontal velocity component: vx = v sin(30°) = (-9.8) sin(30°) = -4.9 m/s. Therefore, the speed at which the rider was traveling off the slope is 4.9 m/s.

2) To determine the horizontal distance the rider travels before striking the ground, we can use the equation x = xi + vit + 0.5at^2, where x is the final horizontal position, xi is the initial horizontal position, vi is the initial velocity in the x-direction, a is the acceleration in the x-direction (which is 0 in this case), and t is the time. We know that the initial horizontal position is 0 m, since the rider starts at the edge of the slope. The initial velocity in the x-direction is also 0 m/s, since the rider is not moving horizontally at the start. The time is still 1s. Therefore, we can solve for x: x =
 

1. How does the height of a projectile affect its range on a race track?

The height of a projectile does not affect its range on a race track. The range of a projectile is determined by its initial velocity and launch angle, not its height.

2. How does air resistance impact the trajectory of a projectile on a race track?

Air resistance can impact the trajectory of a projectile on a race track by slowing it down and causing it to deviate from its ideal path. This is why most race tracks are designed in a way that minimizes air resistance, such as having smooth surfaces and curves.

3. What is the difference between horizontal and vertical velocity in projectile motion on a race track?

The horizontal velocity of a projectile on a race track remains constant throughout its motion, while the vertical velocity changes due to the effects of gravity. The combination of these two velocities determines the overall trajectory of the projectile.

4. How does the angle of launch affect the distance a projectile travels on a race track?

The angle of launch can greatly affect the distance a projectile travels on a race track. The optimal launch angle for maximum distance is 45 degrees, as it allows for the most balanced combination of horizontal and vertical velocities.

5. What is the formula for calculating the range of a projectile on a race track?

The formula for calculating the range of a projectile on a race track is R = (v^2 * sin(2θ)) / g, where R is the range, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. This formula assumes no air resistance and a flat race track surface.

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