Chuyển động ném - Bài tập trung bình

Home Lớp 10 Vật lý 10 - Sách kết nối tri thức Bài 12: Chuyển động némBài tập trung bình

{"common":{"save":0,"post_id":"6428","level":2,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6396","mon_id":"1158790","chapter_id":"1158874","post_id":"6428","question":"M\u1ed9t v\u1eadt \u1edf \u0111\u1ed9 cao h \u0111\u01b0\u1ee3c ném theo ph\u01b0\u01a1ng ngang v\u1edbi t\u1ed1c \u0111\u1ed9 $v_0$<\/span> = 50 m\/s và r\u01a1i ch\u1ea1m \u0111\u1ea5t sau 10 s. L\u1ea5y g = 10m\/$s^2$<\/span>. T\u1ea7m xa c\u1ee7a v\u1eadt là<\/p>","options":["A.<\/strong> 400 m.","B.<\/strong> 400 m.","C. <\/strong>500 m.","D. <\/strong>300 m."],"correct":"3","level":"2","hint":"S\u1eed d\u1ee5ng công th\u1ee9c: L = $v_0$<\/span> . t<\/p>","answer":"L\u1eddi gi\u1ea3i T\u1ea7m bay xa c\u1ee7a v\u1eadt:  L = $v_0$<\/span> . t= 50.10= 500(m) \u0110áp án \u0111úng: C. 500 m<\/strong><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"149","test":"0","date":"2023-02-25 00:27:08","option_type":"txt","len":0},{"id":"6397","mon_id":"1158790","chapter_id":"1158874","post_id":"6428","question":"M\u1ed9t v\u1eadt \u0111\u01b0\u1ee3c ném ngang v\u1edbi v\u1eadn t\u1ed1c $v_0$<\/span> = 30 m\/s, \u1edf \u0111\u1ed9 cao h = 80 m. L\u1ea5y g = 10 m\/$s^2$<\/span>. T\u1ea7m bay xa và v\u1eadn t\u1ed1c c\u1ee7a v\u1eadt khi ch\u1ea1m \u0111\u1ea5t là<\/p>","options":["A. <\/strong>120 m; 50 m\/s.","B.<\/strong> 50 m; 120 m\/s.","C. <\/strong>120 m; 70 m\/s.","D.<\/strong> 70 m; 120 m\/s."],"correct":"1","level":"2","hint":"S\u1eed d\u1ee5ng công th\u1ee9c: L = $v_0$<\/span> . t và $v_cd =\\sqrt{v_0^2+2gh}$<\/span><\/p>","answer":"L\u1eddi gi\u1ea3i Th\u1eddi gian r\u01a1i:  $t= \\sqrt{\\frac{2h}{g}}= \\sqrt{\\frac{2.80}{10}}= 4(s)$<\/span> T\u1ea7m bay xa:  L = $v_0$<\/span> . t= 30.4= 120(m) V\u1eadn t\u1ed1c c\u1ee7a v\u1eadt khi ch\u1ea1m \u0111\u1ea5t:  $v_cd =\\sqrt{v_0^2+2gh}= \\sqrt{30^2+210.80}= 50(m\/s)$<\/span> \u0110áp án \u0111úng: A. 120 m; 50 m\/s.<\/strong><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"149","test":"0","date":"2023-02-25 00:35:10","option_type":"txt","len":2},{"id":"6398","mon_id":"1158790","chapter_id":"1158874","post_id":"6428","question":"T\u1eeb \u0111\u1ed9 cao 15 m so v\u1edbi m\u1eb7t \u0111\u1ea5t, m\u1ed9t v\u1eadt \u0111\u01b0\u1ee3c ném ch\u1ebfch lên v\u1edbi vect\u01a1 v\u1eadn t\u1ed1c \u0111\u1ea7u 20 m\/s h\u1ee3p v\u1edbi ph\u01b0\u01a1ng n\u1eb1m ngang m\u1ed9t góc 300. L\u1ea5y g = 10 m\/$s^2$<\/span>. T\u1ea7m bay xa c\u1ee7a v\u1eadt là<\/p>","options":["A. <\/strong>63 m.","B. <\/strong>52 m.","C. <\/strong>26 m.","D. <\/strong>45 m."],"correct":"2","level":"2","hint":"S\u1eed d\u1ee5ng công th\u1ee9c: $L=v_0.cos\\alpha.t$<\/span><\/p>","answer":"L\u1eddi gi\u1ea3i Ch\u1ecdn g\u1ed1c t\u1ecda \u0111\u1ed9 là v\u1ecb trí ném v\u1eadt, Oy h\u01b0\u1edbng lên. Ch\u1ecdn g\u1ed1c th\u1eddi gian là lúc ném v\u1eadt. Ta có, theo ph\u01b0\u01a1ng Oy:  $y=v_0.sin\\alpha.t - \\frac{gt^2}{2}$<\/span> Khi ch\u1ea1m \u0111\u1ea5t, y = -15 m  $-15= 20.sin30^0.t- \\frac{10t^2}{2}\\implies t=3(s)$<\/span> T\u1ea7m bay xa c\u1ee7a v\u1eadt là:  $L=v_0.cos\\alpha.t= 20.cos30^0.3= 52(m)$<\/span> \u0110áp án \u0111úng: B. 52 m.<\/strong><\/span><\/p>","type":"choose","extra_type":"classic","user_id":"149","test":"0","date":"2023-02-25 00:41:06","option_type":"txt","len":0}]}