Two crickets, Chirpy and Milada, jump from the top of a vertical cliff. Chirpy jumps horizontally and reaches the ground in a time of 2.70 s . Milada jumps with an initial velocity of 81.0 cm/s at an angle of 40.0 degrees above the horizontal.
How far from the base of the cliff will Milada hit the ground?
Take the free fall acceleration to be g = 9.80 m/s^2 .
__________cm
thank you in advance :D
2007-02-06 21:52:22 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Don't have the formula to hand but Chirpy's fall will give you the height of the cliff ie straight drop. When you know the height of the cliff you should be able to work out the angle/height for Milada before she falls and therefore the distance from the base of the cliff.
2007-02-06 22:05:03 · answer #1 · answered by δώδεκα 5 · 0⤊ 0⤋
t = 2.7s
Vi = 0.81m/s
alpha = 40.
Height of the cliff:
h = g . t² / 2
The other one goes up first. Need to find the maximum height:
Hmax = Vi² . sin²alpha / 2g.
This is to be added to the cliff:
Htot = h + Hmax.
Time it takes to go up, then down to reach the same level as the cliff (in an inverted parabola)
t1 = 2 . Vi . sin alpha / g.
Vertical component of the velocity at start - this will be the velocity of the bug when reaching the same level as the cliff, hence, the initial velocity of the continuation of the fall:
Vv = Vi sin alpha.
Then you just have to continue the fall...
The total time it takes x horizontal velocity = distance...
That you can do!
2007-02-07 00:07:49 · answer #2 · answered by just "JR" 7 · 0⤊ 0⤋
Time taken by Chirpy to reach the ground = time taken for Milada
Height of cliff is same for both
g(t^2)/2 = u(sin 40)t - g(t^2)/2 ( u = velocity of Milada )
g(t^2) = u(sin 40)t
u = gt/sin 40
Distance away from cliff's base = u (cos 40) t =
=g(t^2)(cot 40)
Calculate.
2007-02-06 22:31:58 · answer #3 · answered by nayanmange 4 · 0⤊ 0⤋
The height of the cliff can be deduced from the jump of Chirpy
Chirpy has no vertical component of its jump. So vertically, the height of the cliff is
h = 0.5g t^2 = 35.72 m
the jump of Milada. First,milada must reach a maximum and afterreach the ground
When milada jumps the component of gravity in vertical direction is g sin 40° = 9.8 *0.642 =6.3 m/s^2
is initial velocity in the y direction is 0.81*sin 40 °=- 0.521m/s since it jumps inopposite direction pof gravity
the time is 0.81/0.521= 1.56s
the max is hmax = 0.5 *6.3 *(1.56)^2 =7.62 m
So Milada mus fall from 35.72+6.72 = 42.4m
the time for this is 42.4 = 0.5 *6.3 *t^2
you find t =13.2 s
horizontally its speed is 0.81m/s * cos 40° = 0.62m/s
and the distance is 0.62 *13.2 =8.6 m
Hoping it is right!!!
2007-02-07 03:31:33 · answer #4 · answered by maussy 7 · 0⤊ 0⤋
Chirpy takes 2.7 second to fall through a height H.
h = 0.5* 9.8 * 2.7 *2.7 = 35.721 m
Milda’s initial vertical speed is 0.81 sin 40 = 0.5207 m/s.
Displacement = - 35.721 m (take up ward as positive)
acceleration = - 9.8 m/s^2.
- 35.721 = 0.5207 t - 0.5*9.8*t^2.
-7.29 = 0.106 t - t^2
t^2 - 0.106 t -7.29 = 0
Solving
t = 2.753 s or - 2.648s.
Taking the positive time
The horizontal distance covered by Milda is
2.753 x 0.81 x cos 40 = 1.708 m
= 1708 cm.
2007-02-07 02:57:46 · answer #5 · answered by Pearlsawme 7 · 0⤊ 0⤋
please use formulae of projectile motion
height = (intial velocity)^2 * (sin 40)^2 / 2*g
=.81^2 * sin 40 ^2 /2*9.8
2007-02-06 22:08:56 · answer #6 · answered by vyral_143 1 · 0⤊ 0⤋