At a track-and-field meet, the best long jump is measured as 8.00 m. The jumper took off at an angle of 46° to the horizontal.

(a) What was the jumper’s initial speed? (Neglect air resistance.)
1 m/s
(b) If there were another meet on the Moon and the same jumper could attain only half of the initial speed he had on the Earth, what would be the maximum jump there?
2 m
(Air resistance does not have to be neglected in part (b). Why?)

I will be happy to critique your work.

im talking about the ones asked at 7:55 and 7:57...please

7.8 m

To find the answer to part (a), we can use the fact that the initial velocity of the jumper can be split into horizontal and vertical components. The horizontal component of the velocity remains constant throughout the jump, while the vertical component changes due to the effect of gravity.

We know the horizontal component is given by the initial speed multiplied by the cosine of the angle. Therefore, we can write:

Vx = V * cos(θ)

Where Vx is the horizontal component of velocity, V is the initial speed we want to find, and θ is the angle of takeoff.

Next, let's consider the vertical component of the initial velocity. This component only plays a role in determining the time of flight and the peak height reached, but not the distance of the jump. Therefore, we can ignore it and neglect air resistance, as stated in the problem.

Since we are neglecting air resistance and considering only the horizontal component, we can equate the horizontal displacement to the distance of the jump. The horizontal displacement is given by the horizontal component of velocity multiplied by the time of flight, which is twice the time it takes for the jumper to reach the maximum height. So we have:

8.00 m = V * cos(46°) * t

Now we need to find the time of flight. The time taken to reach the maximum height can be found using the vertical component of velocity and the acceleration due to gravity. The vertical component of velocity at the maximum height is zero, and the acceleration due to gravity is -9.8 m/s^2 (taking upward as positive). We can write:

0 = V * sin(θ) - 9.8 * t/2

Notice that the time of flight can cancel out from the two equations, allowing us to solve for V:

8.00 m = V * cos(46°) * [V * sin(46°) / 9.8]

Solving this equation for V will give us the answer to part (a), the jumper’s initial speed.

Now let's move on to part (b). If the same jumper were on the Moon and could only attain half of the initial speed he had on Earth, we need to determine the new maximum jump distance.

Since the horizontal component of velocity is independent of the vertical component, it remains the same on the Moon. Hence, the horizontal component of velocity on the Moon will be half of the initial speed on Earth.

Let's call this new velocity on the Moon V_moon. We know that V_moon is half of the initial speed on Earth:

V_moon = V/2

To find the new maximum jump distance, we can use the same equation as before:

x_moon = V_moon * cos(θ) * t_moon

Since we are on the Moon now, the acceleration due to gravity is different, and we cannot ignore air resistance. On the Moon, the acceleration due to gravity is about one-sixth of its value on Earth, which is approximately 1.62 m/s^2.

Therefore, the equation for the time of flight on the Moon will include the effect of gravity and air resistance:

0 = V_moon * sin(θ) - 1.62 * t_moon/2

So we have two equations involving V_moon and t_moon:

V_moon = V/2

0 = V_moon * sin(θ) - 1.62 * t_moon/2

Now we need to solve these equations to find V_moon and t_moon. Substituting the first equation into the second, we get:

0 = (V/2) * sin(θ) - 1.62 * t_moon/2

Solving this equation for t_moon will give us the time of flight on the Moon. We can then substitute this value of t_moon back into the previous equation for x_moon, replacing t with t_moon, to get the maximum jump distance on the Moon.

That's how we can solve part (b) and determine the maximum jump on the Moon for the same jumper.