The following questions appeared in Kerala Engineering Entrance 2007 question paper:

**(1) Two balls are dropped to the ground from different heights. One ball is dropped two seconds after the other but they both strike the ground at the same time. If the first ball takes 5 s to reach the ground, then the difference in initial heights is (g = 10 ms**^{–2}) ** **

**(a) 20 m**

**(b) 80 m**

**(c) 170 m**

**(d) 40 m**

**(e) 160 m**

The initial height of the first ball is the distance travelled by it in 5 seconds and is given by

*x*_{1} = 0×5 + (½)×10×5^{2}, using the relation, *x = v*_{0}*t +* (½)*at*^{2}

Therefore,* x*_{1} = 125 m.

The time of travel of the second ball is 3 seconds and hence its initial height is given by

*x*_{2} = 0×3 + (½)×10×3^{2} = 45 m.

The difference in initial heights is *x*_{1}– *x*_{2} =125 – 45 = 80 m.

**(2) A ball is thrown vertically upwards with a velocity of 25 ****ms**^{–1} from the top of a tower of height 30 m. How long will it travel before it hits the ground? ** **

**(a) 6 s**

**(b) 5 s**

**(c) 4 s**

**(d) 12 s**

**(e) 10 s**

Let us take all downward vectors positive. The displacement is 30 m and is positive. The acceleration due to gravity, *g *(equal to 10 ms^{–2}) also is positive. But, the velocity of projection is negative (being upwards).

From the equation,** ***x = v*_{0}*t +* (½)*at*^{2}, we have

30 = – 25 t + (½)×10×*t*^{2}

Or, * t*^{2} – 5 t – 6 = 0, from which t = [5±√(25+24)]/ 2 = **+ 6 s** or –1 s.

Since negative time is impossible, the answer is 6 s.

[You could have taken upward quantities as positive if you wanted to, but you would get the same answer].

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