Simple Initial Velocity Problem

[derekfuselier]

An Olympic long jumper leaves the ground at an angle of 18.3 ° and travels through the air for a horizontal distance of 8.86 m before landing. What is the takeoff speed of the jumper?

I'm completely dumbfounded on this one. What I have done is cut the distance in half to create a triangle and then solve all sides in order to find the y. Once I have the way I use the formula ( t = sq rt of 2y/g ) to get the time for the object to fall half way. I multiplied by two to get total time of flight (or what I thought was total time of flight.) Once I have the alleged total time of flight I divide 8.86 by it, giving me the velocity in the x direction. Using cos(18.3) = 8.09 m/s divided by the hypotenuse, or the initial velocity. Solving for the initial velocity left me with 8.52 m/s, which is wrong.

Can anyone offer any help? I can scan my work if needed, I know it's hard to read and think about math like this. Thanks in advance!

 

[PhanthomJay]

An Olympic long jumper leaves the ground at an angle of 18.3 ° and travels through the air for a horizontal distance of 8.86 m before landing. What is the takeoff speed of the jumper?

I'm completely dumbfounded on this one. What I have done is cut the distance in half to create a triangle and then solve all sides in order to find the y. Once I have the way I use the formula ( t = sq rt of 2y/g ) to get the time for the object to fall half way. I multiplied by two to get total time of flight (or what I thought was total time of flight.) Once I have the alleged total time of flight I divide 8.86 by it, giving me the velocity in the x direction. Using cos(18.3) = 8.09 m/s divided by the hypotenuse, or the initial velocity. Solving for the initial velocity left me with 8.52 m/s, which is wrong.

Can anyone offer any help? I can scan my work if needed, I know it's hard to read and think about math like this. Thanks in advance!

Your error is in assuming that the jumper moves in a staight diagonal line to a max height at the apex of the triangle you have described. Since gravity is always acting downward and causing the jumper to continually 'fall' as he jumps, his motion will be parabolic in accordance with projectile motion equations of motion. You should look first in the vertical direction using the appropriate kinematic equation that relates vertical distance with initial vertical speed and time, noting that y = 0 as the jumper hits the ground. Then your second equation comes from looking in the horizontal direction as you have noted. You get 2 equations with 2 unknowns, which can be solved with a bit of trig manipulation. I suggest you do a google search on "projectile range equation" to see an example of a similar problem.

 

[m2003]

I can offer some advice to help you solve this problem. First, let's define some key concepts and equations that will be useful in solving this problem.

1. Initial Velocity: This is the velocity with which the jumper leaves the ground. It is the starting point for our calculations.

2. Angle: The angle of 18.3° is the angle at which the jumper leaves the ground. This will be important in determining the horizontal and vertical components of the initial velocity.

3. Horizontal and Vertical Components: The initial velocity can be broken down into its horizontal and vertical components using trigonometric functions. The horizontal component is given by Vx = V cosθ, where V is the initial velocity and θ is the angle of takeoff. The vertical component is given by Vy = V sinθ.

4. Time of Flight: This is the total amount of time the jumper spends in the air. It can be calculated using the equation t = 2Vy/g, where g is the acceleration due to gravity (9.8 m/s^2).

Now, let's apply these concepts to solve the problem. We know that the horizontal distance traveled by the jumper is 8.86 m. Using the horizontal component equation, we can write:

Vx = 8.86 m

Next, we need to find the vertical component of the initial velocity. Using the vertical component equation, we can write:

Vy = V sinθ

Since we know the angle of takeoff (18.3°), we can plug in this value and solve for Vy:

Vy = V sin18.3°

Now, we can use the equation for time of flight to find the total time the jumper spent in the air:

t = 2Vy/g

Plugging in the value for Vy that we just calculated, we get:

t = 2(V sin18.3°)/9.8

We can now solve for V, the initial velocity:

V = (9.8t)/(2 sin18.3°)

Substituting the value for t that we calculated earlier, we get:

V = (9.8(8.86/2))/(2 sin18.3°)

Simplifying, we get:

V = 8.52 m/s

This is the same answer that you got, so it appears that your calculations were correct. However, it's possible that there was a mistake in one of your