a stone A is thrown vertically upward with speed u and another stone B is thrown vertically downwards with same speed u from a height h. the time taken by stone A and B to reach ground are 16s and 4s respectively. the time taken by another stone to reach the ground after it has been dropped from height h is (g=10m/s
Answers
We will be using the formulae, v = u + at and s = ut + 1/2 * a * t^2. The path of the ball thrown up can be divided into three phases, the going up of the ball (Ball A) from the beginning height to the maximum height (Ph1), then the coming down from max height to beginning height (Ph2), and finally, the going down from beginning height to the surface (earth). Now, as per property, under gravity time taken for Ph1 = time taken for Ph2. The time taken for Ph3 is same as the time taken for Ball B to land, i.e, equal to 4 seconds. So, time for (Ph1 + Ph2) = 16-4 = 12 seconds, and so, time taken for going up or down is 12/2 = 6 seconds.
Now, in Ph1,
V = u + at
0 = u -10 * 6
u = 60
Next, as u is same for both balls,
s = ut + 1/2 * a * t^2
s = 60 * 4 + ½ * 10 * 16
s = 240 + 80 = 320
Finally for the free fall ball,
S = ½*g*t^2
320 = ½ * 10 * t^2
t^2 = 64
t = 8 seconds.
Please comment if anything is not clear. I will try to answer ASAP.
Velocity in Ph2 (VR) is equal to, at height h, will be u.
VR = 0 + 10*6 = 60
So, VR = u.
Answer:
Explanation:the answer is 8 according to the same method as above