phy-力學既問題

Take downward direction as positive, take a=g=10 ms-2

First, for velocity=-v ms-1(as the rock moves upward, opposite to the a)
by v=u+at, when the rock is at its maximum height,
v=0, u=-v,
0=-v+10t
t=v/10 s
by s=ut+1/2 at^2,
maximum height=-v * v/10 + 0.5 * 10 * v^2 /100=-v^2 / 20 m.(negative sign means that the rock is above the point which it is thrown.)
Since the rock moves upwards and downwards, so it takes v/5 s, i.e. t(the t mentioned in the question)=v/5 s

Second, for velocity=-2v ms-1
by v=u+at, when the rock is at its maximum height,
v=0, u=-2v,
0=-2v+10t
t=v/5 s
by s=ut+1/2 * at^2
maximum height=-2v * v/5 + 0.5 * 10 * v^2 /25=-v^2 / 5 m.
so u can see that the distance travelled in air is longer when the velocity is 2v than that when the velocity is v. So the time taken travelling in the air is longer.

So, for velocity is 2v, time taken=2v/5 s=2t...(i.e. it takes double time to travel in the air when the throwing velocity is doubled.)

first,take the upward direction is positive
by v = u + at(1)
v - u / a = t(1)

SECOND
v= u + at(2)
v - 2u/a = t(2)

since s = (v+u)/2 x t

u = 0

s = (v+0)/2 x t

s = v/2 x t

t = 2s/v

since u = 0 and s remains no change

s = (2v+0)/2 x t

s = 2v/2 x t

s = v x t

t = s/v