Question

A ball A is thrown vertically upward from the top of a 24-m-high building with an initial velocity of 7 m/s . At the same instant another ball B is thrown upward from the ground with an initial velocity of 22 m/s .
Determine the height from the ground at which they pass each other.

Determine the time at which they pass each other.

Answer 1

Answer:

The height from ground at which they pass each other is 22.656 m

The time at which they pass each other is 1.6 sec.

Explanation:

For ball A, we have:

height = (h)a

Initial Velocity = Vi = 7 m/s

g = - 9.8 m/s² (for upward motion)

Using 2nd eqn. of motion:

h = Vi t + (1/2)gt²

(h)a = 7t + (1/2)(-9.8)t²

(h)a = 7t - 4.9t²    

this is the height of ball A with reference as the building. Taking ground as reference, we have to add the height of building, that is, 24 m.

(h)a = 7t - 4.9t² + 24  ______ eqn (1)

For ball B, we have:

height = (h)b

Initial Velocity = Vi = 22 m/s

g = - 9.8 m/s² (for upward motion)

Using 2nd eqn. of motion:

h = Vi t + (1/2)gt²

(h)b = 22t + (1/2)(-9.8)t²

(h)b = 22t - 4.9t²    ______ eqn (2)

Now, when the too balls pass each other, there height must be same.

Therefore,

(h)a = (h)b

using eqn (1) and eqn (2):

7t - 4.9t² + 24 = 22t - 4.9t²

22t - 7t = 24

t = 24/15

t = 1.6 sec

Now, for the height, at which they pass each other put t = 6sec in eqn (2)

Therefore,

h = (22)(1.6) - (4.9)(1.6)²

h = 35.2 - 12.544

h = 22.656 m