A helicopter rises from rest on the ground vertically up-wards with a constant acceleration g. A food packet isdropped from the helicopter when it is at a height h. Thetime taken by the packet to reach the ground is close to[g is the accelertion due to gravity] : [5 Sep. 2020 (I)](a) 23ht gæ ö= ç ÷è ø (b) 1.8 ht g=(c) 3.4 ht gæ ö= ç ÷è ø (d) 23ht g=36. A Tennis ball is released from a height h and after freelyfalling on a wooden floor it rebounds and reaches height2h . The velocity versus height of the ball during its motionmay be represented graphically by :(graph are drawn schematically and on not to scale)[4 Sep. 2020 (I)](a)h/2hh v( )v(b)h/2h h v( )v(c) h/2h h v( )v(d) h/2h h v( )v37. A ball is dropped from the top of a 100 m high tower on aplanet. In the last 12 s before hitting the ground, it covers adistance of 19 m. Acceleration due to gravity (in ms–2) nearthe surface on that planet is _______.[NA 8 Jan. 2020 II]38. A body is thrown vertically upwards. Which one of thefollowing graphs correctly represent the velocity vs time?[2017](a) (b)(c) (d)39. Two stones are thrown up simultaneously from the edgeof a cliff 240 m high with initial speed of 10 m/s and 40m/s respectively. Which of the following graph bestrepresents the time variation of relative position of thesecond stone with respect to the first ?www.jeebooks.inP-18 Physics(Assume stones do not rebound after hitting the groundand neglect air resistance, take g = 10 m/ s2) [2015](The figures are schematic and not drawn to scale)(a) (y – y ) m2 12408 12 t(s)(b) (y – y ) m2 12408 12 t(s)(c)(y – y ) m2 12408 12 t(s)t®(d)(y – y ) m2 124012 t(s)40. From a tower of height H, a particle is thrown verticallyupwards with a speed u. The time taken by the particle, tohit the ground, is n times that taken by it to reach thehighest point of its path. The relation between H, u and nis: [2014](a) 2gH = n2u2 (b) gH = (n – 2)2 u2d(c) 2gH = nu2 (n – 2) (d) gH = (n – 2)u241. Consider a rubber ball freely falling from a height h = 4.9 monto a horizontal elastic plate. Assume that the durationof collision is negligible and the collision with the plate istotally elastic.Then the velocity as a function of time and the height asa function of time will be : [2009](a) t+v1vO–v1yht(b)v+v1O–v1t1 2t1 4t1t tyht(c) tt1 2t1Oyht(d)v1vO t tyh42. A parachutist after bailing out falls 50 m without friction. Whenparachute opens, it decelerates at 2 m/s2 . He reaches the groundwith a speed of 3 m/s. At what height, did he bail out ? [2005](a) 182 m (b) 91 m(c) 111 m (d) 293 m43. A ball is released from the top of a tower of height h meters.It takes T seconds to reach the ground. What is the positionof the ball at 3T second [2004](a) 89h meters from the ground(b) 79h meters from the ground(c) 9h meters from the ground(d) 1718h meters from the ground44. From a building two balls A and B are thrown such that A isthrown upwards and B downwards (both vertically). If vAand vB are their respective velocities on reaching theground, then [2002](a) vB > vA(b) vA = vB(c) vA > vB(d) their velocities depend on their masses.www.jeebooks.inMotion in a Straight Line P-191. (b) Given, v = b xor 1/2dx b xdt =or 1/20 0x tx dx bdt- =ò òor1/21/ 2x = 6t or x =2 24b tDifferentiating w. r. t. time, we get2 24dx b tdt´= (t = t)or v =22b t2. (b) Graphs in option (c) position-time and option (a)velocity-position are corresponding to velocity-time graphoption (d) and its distance-time graph is as given below.Hence distance-time graph option (b) is incorrect.timedistance3. (c) Average speed = Total distance travelled xTotal time taken T== xx x2 40 2 60+´ ´= 48 km/h4. (c) We know that, dxv dt=Þ dx = v dtIntegrating,0 0x tdx v dt=ò òor 200( )tx v gt ft dt= + +ò 2 3002 3tgt ftv té ù= + +ê úê úë ûor,2 30 2 3gt ftx v t= + +At t = 1, 0 2 3g fx v= + + .5. (a) v x= a ,Þ dx xdt = a Þ dx dtx = aIntegrating both sides,0 0x tdx dtx = aò ò ; 002 [ ]1xtx té ù = aê úë û2 x tÞ = a2 24x taÞ =6. (a) A BS DCt(in s)v(m/s)420–21 2 3 4 5 6O1 134 3 3OS = + =1 52 3 3SD = - =Distance covered by the body = area of v-t graph= ar (OABS) + ar (SCD)1 13 1 51 4 22 3 2 3æ ö= + ´ + ´ ´ç ÷è ø32 5 37 m3 3 3= + =7. (20) utAB58ODistance travelled = Area of speed-time graph1 5 8 20 m2= ´ ´ =8. (3) Distance X varies with time t as x2 = at2 + 2bt + c2 2 2dxx at bdtÞ = +( )dx dx at bx at bdt dt x+Þ = + Þ =222d x dxx adtdtæ öÞ + =ç ÷è ø2 222dx at ba ad x dt xx xdt+æ ö æ ö- -ç ÷ ç ÷è ø è øÞ = =( )2 23 3ax at b ac bx x2- + -= =Þ a µ x–3 Hence, n = 3www.jeebooks.in