The drawing shows a skateboarder moving at v = 5 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is h = 0.52 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track.

Question

The drawing shows a skateboarder moving at v = 5 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is h = 0.52 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track.

bobpursley I really appreciate all of your help so far, but I am still confused about this question. This is what I've done so far but it is wrong:

mgh= 1/2m x 25 x (sine 48)^2

I solved for h

h = 0.5m x 25 x 0.552/ mg

the m cancels out leaving

h = 0.5 x 25 x 0.552/ 9.81

h = 0.7037

and to find H

H = h - 0.52
H = 0.7037 - 0.52
H = 0.1837

Please help!!!!!!!

Thanks, Mary

It looks right to me. The only question is significant figures. the velocity is given as one significant figure, and h is given as two. Clearly, your answer to four digits is wrong. Technically, the answer ought to be to one digit(.2meters), however, I am not certain your instructor is a nit picker on that. Your work is on target.

Ok. First you take the derivative of the selected quadry, half-end both secular connotations PL-ing the quantum Chancery's over your answer. Simple.

Answer 1

I can understand why you might be confused by this question. Let me walk you through the steps to find the maximum height H.

Step 1: Find the initial vertical velocity component (v_y) of the skateboarder as she leaves the track. To do this, use the given velocity (v = 5 m/s) and the angle of the track (48°). Since the track is slanted upward, the vertical component of velocity can be found using the equation:

v_y = v * sin(angle)

In this case, v_y = 5 m/s * sin(48°) = 3.745 m/s.

Step 2: Determine the time it takes for the skateboarder to reach the highest point (when the vertical velocity component is zero). We can use the kinematic equation:

v_y = v_iy + a_y * t

Since the skateboarder is under the influence of gravity, the vertical acceleration (a_y) is -9.81 m/s^2 (taking downward as the negative direction). The initial vertical velocity (v_iy) is 3.745 m/s. Solving for time (t), we get:

0 = 3.745 m/s - 9.81 m/s^2 * t

Solving for t, we find t ≈ 0.382 seconds.

Step 3: Use the time found in step 2 to calculate the maximum height (H) reached by the skateboarder. The formula to find the maximum height in projectile motion is:

H = y_i + v_iy * t + (1/2) * a_y * t^2

In this case, the initial height (y_i) is 0.52 m and the vertical acceleration (a_y) is -9.81 m/s^2. Substituting the values, we have:

H = 0.52 m + 3.745 m/s * 0.382 s + (1/2) * (-9.81 m/s^2) * (0.382 s)^2

Simplifying the equation gives us the maximum height H.

Step 4: Calculate H using the values obtained:

H ≈ 0.52 m + 1.43 m - 0.69 m ≈ 1.26 m

Therefore, the maximum height H to which the skateboarder rises above the end of the track is approximately 1.26 meters.

Note: Make sure to pay attention to significant figures in your final answer, as your instructor may require specific rounding or significant figure rules.