Question

A boy on a cliff 19.6m high drops a stone. one second later, he throws a second stone after the first. they hit the ground at the same time. with what speed did he throw the second stone

Answer 1

Here,let the time taken by the way stone to travel 19.6 metre be T

S=ut+1/2aT^2
S=1/2gT^2
19.6=1/2×9.8×T^2
T=√(39.2/9.8)
T=2seconds

Second stone is left after 1 sec from the first stone so t=1

S=ut+1/2gt^2
19.6=u×1+4.9×1×1
19.6=u+4.9
u=19.6-4.9
u=14.7m/s

Therefore the second stone was thrown with a velocity of 14.7m/s

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Answer 2

Answer:

Let T be the time taken by the 1st stone to travel

19.6 m.s = u t + 1/2 a t²   19.6   =  1/2 * g * T²

 T =  √(39.2/ 9.8)      

= 2 sec

so, This is the speed for the first stone thrown.  

now for the 2nd stone, we have to find U :

so, 19.6 = u * 1sec + 1/2 g 1²    

therefore, u = 14.7 m/sec

so, This is the speed for the second stone thrown.  

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