Question

L-
A ball A is thrown vertically upwards at 25 ms from a point P. Three seconds later a
second ball B is also thrown vertically upwards from the point P at 25 ms. Taking the
acceleration due to gravity to be 10 ms?, calculate

a. the time for which ball A has been in motion when the balls meet
b. The height above P at which A and B meet.​

Answer 1

Answer:

a. The time for which ball A has been in motion is 4 seconds.

b. The height above P which balls A and B meet is 20m.

Step-by-step explanation:

Parameters given:

Acceleration due to gravity, g = 10m/s

Speed of ball A, u(A) = 25m/s

Speed of ball B, u(B) = 25 m/s

We can solve both problems using the equation of linear motion:

s = ut - 0.5gt^2

Where s = distance travelled

g = acceleration due to gravity

t = time taken

Note: The negative sign is due to the fact that the balls are moving vertically upward, working against gravity.

a. The equation of motion for the first ball can be given as

s(A) = u(A)t - 0.5gt^2

While the equation of motion for the second ball B can be given as

s(B) = u(B)T - 0.5gT^2

Note: t is the time taken for ball A and T is the time taken for ball B.

From the question, after the ball A has been thrown for 3 seconds, ball B is then thrown. This means that

t = T + 3

Fixing that into the E. O. M. for ball A,

s(A) = u(A)×(T + 3) - 0.5×g×(T + 3)^2

= u(A)T + 3u(A) - 5(T^2 + 6T + 9)

= u(A)T + 3u(A) - 5T^2 - 30T - 45

When the two balls meet, that means the distance traveled by both balls are equal, hence,

s = s(A) = s(B)

=> u(A)T + 3u(A) - 5T^2 - 30T - 45 = u(B)T - 5T^2

=> 25T + 75 - 5T^2 - 30T - 45 = 25T - 5T^2

=> 75 - 30T - 45 = 0

=> 30T = 30

T = 1s

t = T + 3

t = 1 + 3

t = 4s

Therefore, the time for which ball A has been in motion is 4s

b. To find the height at which the balls meet i.e. s = s(A) = s(B), we use the equation of motion of ball A:

s(A) = u(A)t - 0.5t^2

s(A) = (25×4) - (0.5×10×4^2)

s(A) = 100 - 80

s(A) = 20m

s = s(A) = 20m

Hence, the balls will meet at a height 20m above point P.

Answer 2

The time for which ball A has been in motion when the balls meet is 4 seconds.

The height above P where  A and B meet is 20 m.

Given,

Velocity of ball A is 25 m/sec.

Velocity of ball B is 25 m/sec.

Acceleration due to gravity (g) is 10.

We can solve both problems using the equation of linear motion. From second equation of motion,

$\rm S=ut+\frac{1}{2} at^2$  Here S is distance traveled, u is the initial velocity and t is time taken.

Since the balls are moving vertically upward, working against gravity so Acceleration a becomes -g, negative sign is due to the fact that the balls are moving vertically upward, working against gravity.

Now, The equation of motion for the first ball will be,

$\rm S(A) = ut - \frac{1}{2} gt^2$

While the equation of motion for the second ball B will be,

$S(B) = uT - \frac{1}{2} gT^2$

Here remember that t is the time taken for ball A and T is the time taken for ball B.

According to the question,  the ball A has been thrown  3 seconds early than ball B is thrown. This means that

$t = T + 3$

putting the value of t in equation of motion for ball  A, we get  

$S(A) = u(T + 3) - \frac{1}{2} g(T + 3)^2$

$S(A)= uT + 3u - \frac{1}{2} \times 10(T^2 + 6T + 9)$

$S(A)= uT + 3u - 5T^2 - 30T - 45$

Since the two balls meet, so distance traveled by both balls are equal, So  

$\rm S(A) = S(B)$

$u(A)T + 3u(A) - 5T^2 - 30T - 45 = u(B)T - \frac{1}{2}\times10 T^2$

$25T + 75 - 5T^2 - 30T - 45 = 25T - 5T^2$

$75 - 30T - 45 = 0$

$30T = 30$

$T = 1$

So,

$t = T + 3$

$t = 1 + 3$

$t = 4$

Therefore, the time for which ball A has been in motion is 4 sec.

Now, To find the height at which the balls meet i.e.$S = S(A) = S(B)$ we use the equation of motion of ball A,

$S(A) = u(A)t - \frac{1}{2} \times 10t^2$

$S(A) = (25\times4)- (5\times4^2)$

$S(A) = 100 - 80$

$S(A) = 20$  

Hence, the balls will meet at a height 20 m above point P.

For more details on Equation of motion follow the link:

https://brainly.com/question/8898885