A long 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.

a.1.14t

b. 12.084m

c.1.6m

Explanation:

a. Given the vertical information as

-We use the to determine the time flight as:

Hence the time of flight is 1.14t

b. Now  given the vertical information as we use the equation to solve for x.

-Note that ax is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 10.6 m/s for vix and 1.1497 s for t , the x can be found to be:

Hence the horizontal distance is 12.084m

c. Finally, we use and the equation with the vertical information to find the y at the peak.

-Substituting and solving we get:

Hence the peak height of the jumper is 1.6m

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6