A long jumper approaches his takeoff board A with a horizontal velocity of 30ft/s. Find the vertical component (vy) of the velocity of his center of gravity at takeoff for him to make the jump if he lands at a spot 22ft from point A. What is the vertical rise of his center of gravity?

So when I started working on it I made separate equations for the x and y components but it didn't turn out right. Is it even necessary to make separate equations?

Get his time in the air, T, from the horizontal distance and velocity component.

T = 22/30 = 0.7333 s

The vertical velocity component must decelerate to zero in half that time (when maximum elevation is reached)

Vy = g*(T/2)= 3.59 ft/s

The vertical rise of the CG is given by
(g/2)*(T/2)^2 = H
or (Vy/2)*(T/2)
They give the same result.

I see how this would be correct but the book says the answer is 11.81ft/s and the vertical rise is 2.16 ft

His methods for finding the answer is correct. The only problem is that he uses 9.81 for acceleration by gravity and in this particular problem all dimensions are in ft. So use 32.2 ft/sec and you will find the right answer.

Yes, it is necessary to separate the equations for the x and y components in this problem. This is because the motion of the jumper can be broken down into two independent motions: one in the horizontal direction (x-component) and one in the vertical direction (y-component). By considering the two components separately, you can accurately calculate the required values.

To solve this problem, you can use the following equations:

1. For the x-component of velocity (vx): vx = 30 ft/s (given)

2. For the y-component of velocity (vy): vy = ?

3. For the horizontal displacement (dx): dx = 22 ft (given)

4. For the vertical displacement (dy): dy = ?

The jumper's motion can be described using the following kinematic equation:

dy = vy * t + (1/2)*a*t^2,

where t is the time of flight and a is the acceleration due to gravity (approximately 32 ft/s^2).

Since the horizontal velocity (vx) remains constant and the displacement (dx) is also known, the time of flight (t) can be calculated using the equation:

t = dx / vx.

Substituting the given values:

t = 22 ft / 30 ft/s = 0.7333 s.

Now, we can substitute this value of time into the equation for the vertical displacement (dy):

dy = vy * t + (1/2)*a*t^2.

Since the jumper is taking off and landing at the same horizontal level (which means there is no change in vertical position), dy = 0.

Using this information, the equation becomes:

0 = vy * t + (1/2)*a*t^2.

Now, substitute the known values:

0 = vy * 0.7333 s + (1/2)*32 ft/s^2 * (0.7333 s)^2

Solving this equation will give you the value for vy, which is the vertical component of the jumper's velocity at takeoff. Then, you can use this value to determine the vertical rise of his center of gravity.