Misalin motsi a tsaye sama

Misalai 3 na tambayoyin motsi a tsaye sama

1. Ana jefa ƙwallon a tsaye sama da saurin farko na 20 m/s. A ƙayyade matsakaicin tsayin ƙwallon. g = 10 m/s²
Tattaunawa
A cikin motsi na tsaye sama, lokacin da abu ya motsa sama, yana raguwa, kuma idan ya koma ƙasa, yana ƙaruwa. Saboda haka, motsi na tsaye sama shima misali ne na GLBB.
Tsarin GLBB :
v_t = v_o + a
s = v_o t + ½ a²
v_t² = v_o² + 2 axles
Tsarin GLBB da ke sama an gyara shi kuma an daidaita shi zuwa yanayin motsi na tsaye sama, tare da bayanai da yawa.

Tsarin Motsi Mai Tsaye Zuwa Sama :
v_t = v_o + gt
h = v_o t + ½ gt²
v_t² = v_o² + 2 gh
Bayani: v_t = gudun ƙarshe, v_o = saurin farko, g = hanzari saboda nauyi, t = tazara tsakanin lokaci, h = tsayi.
Catatan :
Na farko, wajen magance matsalolin motsi a tsaye sama, adadin vektor da aka nuna sama ana ba shi alama mai kyau, adadin vektor da aka nuna ƙasa ana ba shi alama mara kyau.
Na biyu, idan matsayin ƙarshe na abu ya fi matsayin farko (matsayin farko shine wurin tunani) to canjin (h) na abu yana da kyau. Akasin haka, idan matsayin ƙarshe ya kasance ƙasa da matsayin farko to canjin abu yana da kyau.
Na uku, a matsakaicin tsayi, abu yana nan na ɗan lokaci kafin ya canza alkibla, saboda haka saurin abu = 0.

An san cewa:
v_o = 20 m/s (tabbatacce saboda alkiblar saurin farko yana sama = an jefa abin sama), g = – 10 m/s² (mara kyau saboda alkiblar hanzarin nauyi koyaushe ƙasa take, v_t = 0 (a matsakaicin tsayi, abu yana hutawa na ɗan lokaci)
An tambaya:
Tsawon mafi girma (h)?
Amsa:

2. Ana jefa marmara a tsaye daga gini mai nisan mita 100 daga ƙasa tare da saurin farko na mita 20/s. A ƙayyade (a) lokacin da ake buƙata don isa ƙasa (b) saurin marmarar lokacin da ta isa ƙasa. g = 10 m/s²
Tattaunawa
An san cewa:
h = mita -100
v_o = 20m/s
g = -10 m/s²
An tambaya:
(a) tazara tsakanin lokaci (t)
(b) gudun ƙarshe (v)_t)
Amsa:
(a) tazara tsakanin lokaci (t)
Idan aka ba da h = mita -100 (mara kyau saboda matsayin ƙarshe na marmara yana ƙasa da matsayin farko na marmara), v_o = 20 m/s (tabbatacce saboda alkiblar saurin farko yana sama ko kuma alkiblar motsi na farko yana sama), g = -10 m/s² (mara kyau saboda alkiblar saurin nauyi tana ƙasa).

Lokaci ba zai iya samun ƙimar da ba ta da kyau ba, don haka ana amfani da t.₂ = daƙiƙa 6,9.

(b) Saurin ƙarshe
Da aka ba da h, v_o kuma g, ya tambayi v_t, don haka yi amfani da dabara ta uku.

3. Ana jefa ƙwallon A a tsaye sama da gudun 10 ms^-1Daƙiƙa ɗaya bayan haka, an jefa ƙwallon B a tsaye sama a kan hanya ɗaya da gudun 25 ms.^-1Tsawon da ƙwallon B ta kai lokacin da ta haɗu da ƙwallon A shine…

A. 0,20 m

B. 4,80 m

C. mita 5,00

D. 5,20 m

E. 31,25 m

Tattaunawa

A cikin magance matsalar motsi a tsaye sama, adadin vector da aka nuna sama ana ba su alama mai kyau, adadin vector da aka nuna ƙasa ana ba su alama mara kyau.

An san cewa:

Saurin farko (v)_o) ƙwallon A = 10 m/s

Tazarar lokaci (t) da ƙwallon A ke cikin iska = x

Saurin farko (v)_o) ƙwallon B = 25 m/s

Tazarar lokaci (t) da ƙwallon B ke cikin iska = x - 1

Saurin gudu saboda nauyi (g) = -10 m/s² (alkiblar saurin nauyi tana ƙasa don haka tana da alama mara kyau)

An tambaya: Tsawon da ball B ya kai lokacin da ya hadu da ball A (h)

Adadin da ke akwai shine saurin farko (v)_o), hanzarin nauyi (g), tsayi (h) da tazara ta lokaci (t) don haka dabarar da aka yi amfani da ita ita ce:

h = v_o t + ½ gt²

Domin a haɗu, tsayin ƙwallo biyu dole ne ya zama iri ɗaya.

h_A = h_B

v_o t + ½ gt² = v_o t + ½ gt²

10x + ½ (-10) x² = 25 (x-1) + ½ (-10) (x-1)²

10x - 5x² = 25 (x-1) – 5 (x-1)²

10x - 5x² = 25x – 25 – 5 (x²-2x+1)

10x - 5x² = 25x – 25 – 5x² 10x-5 ku

10x - 5x² - 25x + 25 + 5x² - 10x + 5 = 0

- 5x² + 5x ku² + 10x – 25x – 10x + 25 + 5 = 0

10x – 25x – 10x + 25 + 5 = 0

- 25x + 25 + 5 = 0

- 25x + 30 = 0

- 25x = – 30

x = -30/-25

x = daƙiƙa 1,2

Tazarar lokacin da ƙwallon A ke cikin iska kafin ta haɗu da ƙwallon B = daƙiƙa 1,2.

Tazarar lokacin da ƙwallon B ke cikin iska kafin ta haɗu da ƙwallon A = daƙiƙa 1,2 – daƙiƙa 1 = daƙiƙa 0,2.

Tsawon da ƙwallo A ta kai lokacin da ta haɗu da ƙwallo B (h):

h = v_o t + ½ gt² = (10)(1,2) + 1/2 (-10)(1,2)² = 12 – 5(1,44) = 12 – 7,2 = 4,8 mita

Tsawon da ball B ya kai lokacin da ya hadu da ball A (h):

h = v_o t + ½ gt² = (25)(0,2) + 1/2 (-10)(0,2)² = 5 – 5(0,04) = 5 – 0,2 = 4,8 mita

Amsar da ta dace ita ce B.

Tambayoyi game da motsi a tsaye sama

1. Ana jefa ƙwallon a tsaye sama da saurin farko na 10 m/s. A ƙayyade matsakaicin tsayin ƙwallon. g = 10 m/s²
Amsa:
h = mita 5
2. Ana jefa marmara a tsaye daga gini mai nisan mita 50 daga ƙasa tare da saurin farko na mita 5/s. A ƙayyade (a) lokacin da ake buƙata don isa ƙasa (b) saurin marmarar lokacin da ta isa ƙasa. g = 10 m/s²
Amsa:
(a) t = daƙiƙa 3,7 (b) v_t = 32m/s

[Turanci: Motsi sama da ƙasa a cikin faɗuwar 'yanci - matsaloli da mafita]