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Question

From the top of a 11 m high tower a stone is projected with speed 10 m/s, at an angle of 37 as shown in the figure. (Take g=10 m/s2)


Match the quantities in column-I to their values in column-II

Column-I Column-II
(i) Speed after 2 s p. 85
(ii)Time of flight q. 12.8
(iii)Range r. 115
(iv) The maximum height attained by the stone s. 885
(v) Speed just before striking the ground. t. 260

A
(i) - r, (ii) - s, (iii) - p, (iv) - q, (v) - t
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B
(i) - p, (ii) - t, (iii) - s, (iv) - q, (v) - r
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C
(i) - t, (ii) - r, (iii) - s, (iv) - q, (v) - p
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D
(i) - p, (ii) - r, (iii) - s, (iv) - q, (v) - t
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Solution

The correct option is C (i) - t, (ii) - r, (iii) - s, (iv) - q, (v) - p
a) Initial velocity in horizontal direction =10cos37=8 m/s
Initial velocity in vertical direction =10sin37=6 m/s
Velocity after 2 seconds
v=vx^i+vy^j=8^i+(uy+ayt)^j=8^i+(610×2)^j=8^i14^j
Speed (magnitude of velocity) =82+142=260 m/s

b) Using second equation of motion, Sy=uyt+12ayt2
11=6×t+12×(10)t2
5t26t11=0(t+1)(5t11)=0
t=115 sec (taking positive value)
c) Range =ux×t=8×115=885 m

d) Maximum height above the level of projection, h=u2y2g=622×10=1.8 m

Maximum height above ground =11+1.8=12.8 m

e) Speed just before striking the ground v=v2x+v2y=u2x+u2y+2gh=(8)2+(6)2+2×(10)×(11)=100+2×10×11
v=85 m/s

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