# Acceleration due to gravity of g = 6.794 m/s2 and want to jump safely into the water below after leaping from a 9.00 meter high-cliff. ?

You are on a planet that has an acceleration due to gravity of g = 6.794 m/s2 and want to jump safely into the water below after leaping from a 9.00 meter high-cliff. If there is a ledge right at the bottom that is 1.75 m wide, and assuming your initial velocity is directed entirely along the x-axis, how fast do you need to be running in order to land in the water?

Relevance
• y(t) = ½gt² + v₀t  + y₀, v=at and d=vt are the basic formula you need to know.

g = 6.794 m/s²

First: get time of fall

Then: get velocity of distance/time • You need to know the time t to impact. That is the time he has at horizontal speed Vx to travel 1.75m

Vx*t = 1.75

We know 9m = ½at² so t = √(2*9/a) = √(18/6.794) = 1.628

Vx = 1.75/t = 1.75/1.628 = 1.075m/s or 1.08m/s with 3 s.d.

• Tfall = 1.628s

V=1.628/1.75=1.075m/s.

• Anonymous
1 month ago

Time in air = sqrt.(2h/g) = sqrt.(18/6.794) = 1.628 secs (3 decimal places).

To clear the ledge, in 1.628 secs. you must move (say) 2 metres horizontally.You must start with a horizontal speed of (2/1.628) = 1.228 m/sec.If you really must use 1.75 metres for ledge width, the horizontal V needed is (1.75/1.628) = 1.075 metres/ sec.

• falling time t = √2h/g = √2*9/6.794 = 1.63 sec

V = d/t = 1.75/1.63 = 1.075 m/sec (1.08 with just 3 sign. figures)

• Time of fall is T = sqrt(2S/g); so to go beyond that ledge D = 1.75 = Ux T which means Ux = D/T = 1.75/sqrt(2*9/6.794) = 1.08 m/s  ANS.

• g = planet's surface gravity = 6.794 m/s²

y = height of cliff = 9.00 m

w = width of ledge = 1.75 m

v0y = initial vertical velocity = 0 m/s

vy = final vertical velocity = to be determined

v0x = initial horizontal velocity = to be determined

(vy)² = (v0y)² + 2gy

(vy)² = (0 m/s)² + 2(6.794 m/s²)(9.00 m)

(vy)² = 122.292 m²/s²

vy = 11.05857134 m/s

vy = v0y + gt

vy - v0y = gt

(vy - v0y) / g = t

t = (vy - v0y) / g

t = (11.05857 m/s - 0 m/s) / 6.794 m/s²

t = 1.627696694 s

v0x > w / t

v0x > 1.75 / 1.627697 s

v0x > 1.07513888 m/s NECESSARY INITIAL VELOCITY

• Barely missing the 1.75 m ledge... Looks like I'm the first to answer...

If you run/jump horizontally  [ no vertical velocity ], then   h = 0.5 g t^2

h = 9,  g = 6.794    so  t =  1.628 sec until you hit the water.

Horiz   x = Vt ,  V = horiz  velocity.

x = 1.75 m,  and  t   from above,  so  V = 1.075 m/sec  to Barely miss the overhang  below.