Question

a ball is thrown with a velocity of 4m/s from a 40m tower at the same time another ball is thrown from the ground with the speed of 24m/s.Find the time where they meet and the position??please tell the correct answer​

Answer 1

Answer:

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we have u1 = 40m/s and u2 = 24m/s

Clearly, since u2 < u1, the 2 balls will not meet while both are ascending, but they could meet when ball #1 is descending.

So s1 = 40 + ut + ½at^2 = 40 + 40t – 4.095t^2

And s2 = 24t – 4.095t^2

Now, s1 = s2 (when they meet)

40 + 40t – 4.095t^2 = 24t – 4.905t^2

-16t = 40, so t = -2.5

The 2 balls are not going to meet, because the total flight time for ball #1 is 9.5s and for ball #2 is 4.9s. So, the flight of ball #2 will be over, before ball #1 descends to an altitude that could intersect with the path of ball #2.

HOWEVER,

If the ball is thrown DOWNWARDS from the top of the tower, then this question will have a numerical answer.

In this case,

s1 = 40 + ut + ½at^2 = 40 - 40t – 4.095t^2

And s2 = 24t – 4.095t^2

Now, s1 = s2 (when they meet)

40 – 40t – 4.095t^2 = 24t – 4.905t^2

40 – 64t = 0

t = 0.625s

s2 = 24(0.625) – 4.905(0.625^2) = 13.08m

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