Augustine continues to run around. He gets to the park and has the overwhelming urge to jump off a cliff and try to fly. He does so at an angle of 50 degrees to the horizontal while running at 14.5m/s. With his arms flapping, he rises in the air.

a) what is the highest point from the top of the cliff that he reaches before the laws of physics catch up with him and start bringing him back down to Earth?
b)What is his velocity at that point?
c)If the cliff is 10m high, how long is he in the air before he comes crashing down to the nice, soft, sandy beach below the cliff?
d)what is his final velocity?
e)what is his final horizontal displacement from the cliff?

Vo = 14.5m/s @ 50 Deg.

Xo = 14.5cos50 = 9.32 m/s.
Yo = 14.5sin50 = 11.1 m/s.

a. Yf^2 = Yo^2 + 2g*h,
h = (Yf^2 - Yo^2) / 2g,
h = (0 - ( 11.1)^2) / -19.6 = 6.29 m.

b. V = Vf = 0 = Final Velocity.

c. d = Vo*t + 0.5g*t^2 = 6.29+10.
0 + 4.9t^2 = 16.29,
t^2 = 3.32,
t = Tf = 1.82 sw. = Fall time.

d. Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*16.29 = 319.28,
Vf = 17.9 m/s.

e. Tr = (Vf - Vo) / g,
Tr = (0 - 11.1) / -9.8 = 1.13 s. = Rise
time or time to reach max ht.

Tr + Tf = Time in flight.
Dx = Xo * (Tr+Tf),
Dx = 9.32 m/s * (1.13+1.82)s = 27.5 m.

post it.

To solve this problem, we can use the principles of projectile motion. Let's break down each part of the question step by step:

a) What is the highest point from the top of the cliff that Augustine reaches before coming back down?

To find the highest point, we need to calculate the vertical component of his initial velocity. The initial vertical velocity can be calculated using the formula:

Vy = V * sin(θ)

where Vy is the vertical component of the initial velocity, V is the initial velocity, and θ is the launch angle.

In this case, Vy = 14.5 m/s * sin(50°). Calculate this value to find the vertical component of Augustine's initial velocity.

b) What is his velocity at the highest point?

At the highest point of the trajectory, the vertical velocity becomes zero. However, the horizontal velocity component remains constant throughout the motion.

c) If the cliff is 10m high, how long is Augustine in the air before coming down?

To find the time of flight, we can use the formula:

Time = (2 * Vy) / g

Where g is the acceleration due to gravity (approximately 9.8 m/s^2). Plug in the value of Vy calculated in part a) to find the time of flight.

d) What is Augustine's final velocity?

At the maximum height, Augustine's final velocity will have both vertical and horizontal components. However, the magnitude of the horizontal component will remain constant. The vertical velocity component at the top will be the same magnitude but in the opposite direction as the initial vertical velocity. Thus, projecting the velocity components back together, the final velocity can be found using the formula:

V_final = sqrt((Vx^2) + (-Vy^2))

Where Vx is the horizontal component of the initial velocity, and -Vy is the negative vertical component at the highest point.

e) What is his final horizontal displacement from the cliff?

To find the horizontal displacement, we can use the formula:

Horizontal Displacement = Vx * Time

Where Vx is the horizontal component of the initial velocity, and Time is the total time of flight calculated in part c).

By following these steps and calculations, you will be able to answer each part of the question.