Question

1) An object is dropped from the top of tower of hight 156.8m and at the same time another object is thrown vertically upward with the velocity of 77.1 m/s from the feetof tower,when and where the object meet​

Answer 1

Answer:

The objects meet after 2.033722438s, at a height 136.5334679m from the ground.

Explanation:

Call the object being dropped A and the object being thrown B.

When the objects meet, the time ('t') for both objects will be equal, and the sum of their distances will be equal to 156.8m.

Figure out the distance for A in terms of time, t:

s = ut + (1/2)at²

= 0t +(1/2)(9.8)t²

=4.9t²

Figure out the distance for B in terms of time, t:

s = ut + (1/2)at²

=(77.1)t + (1/2)(-9.8)t²

=77.1t - 4.9t²

We now adding the distances gives us 156.8, so can write:

(4.9t²) + (77.1t - 4.9t²) = 156.8

77.1t = 156.8

∴ t = 2.033722438s

To find the distances, plug in this value for t:

For A: 4.9t² → 20.26653209m

For B: 77.1t - 4.9t² → 136.5334679m