Question

Remember that the velocity in the Y
(vertical) direction (Vy) is equal to the
V*sin(angle). Remember also that HALF
WAY the vertical velocity is 0.
A long jumper leaves the ground at an angle
of 23.1 to the horizontal and at a speed of
9.71 m/s.
How far does he jump? The acceleration
due to gravity is 9.8 m/s
Answer in units of m.

What maximum height does he reach?
Answer in units of m.

Answer 1

u = 9.71 cos 23.1 = 8.93
Vi = 9.71 sin 23.1 = 3.81

Vy = Vi - g t = 3.81 - 9.8 t
at top Vy = 0
then at top 9.8 t = 3.81
t = .389 second at top (vertex of parabolic path)
falls for the same amount of time
so .389 * 2 = .777 second in air
distance = u t = 8.93 * .777 = 6.94 meters long jump
y = Yi + Vi t - 4.9 t^2 where t is time up = .389 s
y = 0 + 3.81 (.389) - 4.9(.389)^2
= 1.48 - .741 = .739 meter