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?
- JimLv 73 weeks ago
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
- oldschoolLv 73 weeks ago
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.
- PhilomelLv 73 weeks ago
Tfall = 1.628s
- Anonymous3 weeks 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.
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- oubaasLv 74 weeks ago
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)
- oldprofLv 74 weeks ago
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.
- ?Lv 74 weeks ago
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
- MathguyLv 54 weeks ago
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.