For the long jumper that leaves the ground with an initial velocity of 12 m/s, we have:
a. The time of flight is 1.15 s.
b. The horizontal distance is 12.2 m.
c. The peak height of the long-jumper is 1.62 m.
a. We can find the time when the jumper reaches the maximum height as follows:
[tex] v_{f_{y}} = v_{i_{y}} - gt [/tex]
Where:
[tex] v_{f_{y}} [/tex]: is the final velocity in the y-direction = 0 (at the maximum height)
[tex] v_{i_{y}} [/tex]: is the initial velocity in the y-direction = vsin(θ)
g: is the acceleration due to gravity = 9.81 m/s²
θ: is the angle (above the horizontal) = 28°
t: is the time
Solving equation (1) for "t", we have:
[tex] t = \frac{v_{i_{y}}}{g} [/tex]
Since the above time is the rise time (to maximum height), the flight time is:
[tex] t_{f} = 2t = 2\frac{v_{i_{y}}}{g} = \frac{2*12 m/s*sin(28)}{9.81 m/s^{2}} = 1.15 s [/tex]
Hence, the time of flight is 1.15 s.
b. The horizontal distance can be calculated as follows:
[tex] x_{f} = x_{i} + v_{i_{x}}t_{f} + \frac{1}{2}at_{f}^{2} [/tex]
Where:
[tex] x_{f} [/tex]: is the final horizontal distance =?
[tex] x_{i} [/tex]: is the initial horizontal distance = 0
[tex]v_{i_{x}}[/tex]: is the initial velocity in the x-direction = vcos(θ)
a: is the acceleration = 0 (in the x-direction)
Then, the horizontal distance is:
[tex] x_{f} = v_{i_{x}}t_{f} = 12 m/s*cos(28)*1.15 s = 12.2 m [/tex]
Therefore, the horizontal distance is 12.2 m.
c. The peak height (maximum height) of the long-jumper is:
[tex] v_{f_{y}}^{2} = v_{i_{y}}^{2} - 2gh [/tex]
Where:
h: is the maximum height =?
So, the peak height is:
[tex] h = \frac{v_{i_{y}}^{2} - v_{f_{y}}^{2}}{2g} = \frac{(12 m/s*sin(28))^{2}}{2*9.81 m/s^{2}} = 1.62 m [/tex]
Therefore, the peak height of the long-jumper is 1.62 m.
Learn more about maximum height here:
- https://brainly.com/question/6261898?referrer=searchResults
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I hope it helps you!
