Question

From the top of a tower 100m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with velocity of 25m/s. Find when and where the balls will meet? Take g = 9.8ms-2

Answer 1

Explanation:

Form the top of a tower 100 m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velcoity of 25 ms^(-1). ... Hence, the two balls will meet after 4seconds at a distance 78.4m below the top

Answer 2

If the distance covered by ball 2 is 100 - x, then the distance covered by ball 1 will be x, Since the total hieght is 100m.

Since acceleration is constant, Therefore we can apply equations of motion for both balls (considering Positive Downwards).

For ball 1:

Initial velocity, u = 0

Acceleration, a = g = 10 m/s

Displacement, s = x

Now, s = ut + 1/2[at^2]

=> x = 0* t + 1/2 * 10 * t^2

=> x = 5t^2      ----eq 1

For ball 2:

Initial velocity, u = -25m/s  (v is upwards so its -ve, By sign convention)

Acceleration, a = g = -10 m/s ( ball is going upwards so its -ve, By sign convention)

Displacement, s = -(100 - x) = x - 100

Now, s = ut + 1/2[at^2]

=>  x - 100 = (-25)* t + 1/2 * (-10) * t^2

=>  x - 100 =  (-25)* t + 1/2 * (-10) * t^2    

=>  x =  -25t - 5t^2  + 100   ----eq 2

Now from equation (1) and (2), we get,

-25 t - 5t^2  + 100  = 5t^2

=> 25t = 100

=> t = 4 seconds

Now, x = 5t^2      ---- from eq 1

(Putting value of t = 4s)

x = 5 * 4^2

x = 16 * 5

x = 80 meter

Hence, the balls will meet at (100 - 80) = 20m above the ground after 4 seconds.