Along jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

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:

Where:

: is the final velocity in the y-direction = 0 (at the maximum height)

: 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:

Since the above time is the rise time (to maximum height), the flight time is:

Hence, the time of flight is 1.15 s.

b. The horizontal distance can be calculated as follows:

Where:

: is the final horizontal distance =?

: is the initial horizontal distance = 0

: is the initial velocity in the x-direction = vcos(θ)

a: is the acceleration = 0 (in the x-direction)

Then, the horizontal distance is:

Therefore, the horizontal distance is 12.2 m.

c. The peak height (maximum height) of the long-jumper is:

Where:

h: is the maximum height =?

So, the peak height is:

Therefore, the peak height of the long-jumper is 1.62 m.

Learn more about maximum height here:

I hope it helps you!

flight time: 1.5s, horizontal distance: 12.2m, peak height: 1.6m

Explanation:

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.