Solve this sum

Solve this sum Level 3 erhe (x) variation of particle ntnc — _ 15t2 .30 find the (t) given by expression x — posltlon of the body at t = s and = s. Is this parttcle moving With urutOrrn acceleranon? Explain. horizontal range of a bullet fired wath of prejectlon 450 to the horizontal 360 nw If It fired from a lorry moving In the direction of bullet With the uniförrn velocity 18 km hr¯' and Wlth same elcwation, what is the new range horizontal distance traveled by the bullet? (Take g = 10 m s¯2). 63. The graph of vertical displacement (y) versus horizon- tal displacement (x) ofa projectile IS as shown below. (m) 40 10 20 (m) From the values given in the graph, find the time at which the projectile has the displacements, as indi- cated in the graph. 64. A body moving linearly with uniform acceleration covers distances p, q, r and s after successive intervals of time, t s. Find the ratio of 65. A small steel ball is of 2.5 m into a tall glycerin jar. «hits the surface of

Dear Student

62. The new range will be the total of range and additional distance covered by lorry in time of flight T.
So we need to find out first time of flight and then additional distance covered by lorry in that time.
Range R = v2Sin 2θgv2=RgSin 2θ=360 x 10Sin 90=3600v =60 m/sTime of flight T =2v Sinθg=2×60×Sin 4510=8.48sHorizontal distance covered by lorry x = vxTvx= 18 km/h = 5m/sx = vxT= 5×8.48= 42.42 m
So new range will be
360 m +42.42 m = 402.42 m

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