Question

the acceleration ac x=1.9 m Is maximum height from the ground (m) A particle moves along a straight line with acceleration a = 1.33* V m/s where V is the velocity (m/s) the velocity at t=1.9 sis knowing that:-0.X-0.V0-9.3 m/s Select one: Select one O A. 283.33 CA. 116.60 Q4 Q8 CB. 143.32 Select one: A. 21.72 OB. 14.93 C. 19.45 Q1 O 0.487.34 O c.0.00 OD. 147.33 C C. 80.97 CD 170.04 O E. 204.00 OD. 17.19 C E. 134.42 O E. 11.31 It is observed that the time for the ball to strike the ground at B is 8.1 s.a- y-component of velocity at B (m/s) Vo the acceleration at l=1.9 sis Select one: CA-48.60 a 1.2 m B CB.-17.53 Select orie: OA 343.78 Q2 Q9 0 3. 152.79 30 m O C.-26.29 CD0.00 E.-39.84 VON (m/s) time of maximum height (5) O C.477.47 0.248.28 0.391.52 the velocity at x-1.9 mis Select one: OA. 84.65 X-component of acceleration at B (m/s) B. 35.55 Se ect one: A.55.79 Selecione O A.5.89 C. 0.00 OD 52.48 E.44.02 B. 29.71 C. 48.85 Select one: O A.-17.53 OB. 2.42 OC. 4.92 Select one: OA4.94 O B.-26.29 Q3 OB 20.15 D. 0.00 E 30.58 Q6 OD.3.55 O C.-48.60 Q5 Q10 Q09.50 OE.4.03 Q7 OD. 0.00 OD 12.55 O E. 10.35 O E.-39.84

Answer 1

gives 9, a=0.33 Ju m/s2 Vo = 9.3 m/s t=o X=0 dv 0.33 Sv dt dv = 0.33 SV dt 33fd dv - 0.33ldt vo V + V 0.33 33 [+] kt ytl Vo afv-vos] 0-33t at t= 1.9 sec. do - 0.33 (3.9) + Sais 11,3) mis

at t=1.9 acc 92) VE 11:3) at t=109. 0.33 11:31 q a 1.109819 m2 > 1.9m. at velocity velocity at Sax 8 qds andr 0.330 ds = v di du 0:33|1s = Ss tc 0.33 (5)= 2035 21 f=2= at v=vo s l2 312 2 (9-3) a ca -2 Vo so/m 3 at SEX=19 mln 312 2 (9-3) 0.3361.9) a olm 2 (0.2 3 3 ve 13 * 0-33(18) + (9-3)? (4-3)? V = 9.504 m/s

& IN 2-1.am acceleration at at x=109m V= 9.50H m/s a = 0:33 ST 1.01736 m/s2 In this accelerations options 2 are 9. getting but the not matching. So please cherk As our poresses cossect thinks options are given wrong. I