u = V cos 29.7 the whole time
x = u t
Vi = V sin 29.7 up velocity at start
v = Vi - 9.81 t
h = Vi t - (9.81/2) t^2
at top, v =0
so
t = Vi/9.81
h = .54 = Vi (Vi/9.81) - (9.81/2)( Vi^2/9.81^2)
.54 = (1/2)Vi^2/9,81
1.08*9.81 = Vi^2
etc
get Vi
get t
get u =2.95/t
get V = u^2+Vi^2
Salmon often jump waterfalls to reach their
breeding grounds.
Starting downstream, 2.95 m away from a
waterfall 0.54 m in height, at what minimum
speed must a salmon jumping at an angle of
29.7
◦
leave the water to continue upstream?
The acceleration due to gravity is 9.81 m/s
2
.
Answer in units of m/s.
6 answers
ah
V = sqrt (u^2 + Vi^2)
V = sqrt (u^2 + Vi^2)
sorry damon but what is the variable u is in this example
u = horizontal speed the whole time (no horizontal force so no horizontal acceleration
starting from 2.95 m downstream, a line at an angle of 29.7º
... passes over the waterfall 1.68 m above the stream
this means the salmon's trajectory can drop (1.68 - .54) m during the jump
1.14 = 1/2 g t^2 ... 2.28 / g = t^2 ... t = .48 s
the horizontal component of the launch velocity is ... 2.95 m / .48s
the launch velocity is ... (horizontal component) / cos(29.7º)
... passes over the waterfall 1.68 m above the stream
this means the salmon's trajectory can drop (1.68 - .54) m during the jump
1.14 = 1/2 g t^2 ... 2.28 / g = t^2 ... t = .48 s
the horizontal component of the launch velocity is ... 2.95 m / .48s
the launch velocity is ... (horizontal component) / cos(29.7º)
I misread it. R Scott did it right.
0 answers