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??​

Answer 1

Answer:

A ball is thrown from a tower of height 40m with a velocity of 40m/s and another ball is thrown with a velocity of 24m/s from the bottom of the tower. When and where will the 2 bodies meet with respect to the first ball?

A ball is thrown from a tower of height 40m with a velocity of 40m/s and another ball is thrown with a velocity of 24m/s from the bottom of the tower. When and where will the 2 bodies meet with respect to the first ball?

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

Answer 2

Answer:

A ball is thrown from a tower of height 40m with a velocity of 40m/s and another ball is thrown with a velocity of 24m/s from the bottom of the tower. When and where will the 2 bodies meet with respect to the first ball?

A ball is thrown from a tower of height 40m with a velocity of 40m/s and another ball is thrown with a velocity of 24m/s from the bottom of the tower. When and where will the 2 bodies meet with respect to the first ball?

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