Question

A ball vertically thrown upwards with a speed of 19.6m/s from the top of a tower returns to earth in 6 s . find the height of the tower..without using the equations of motion..​

Answer 1

Answer:

Very Very Long Answer Ahead

Explanation:

Acceleration is basically rate of change of vel - m/s^2

m/second ^2 means that for every passing second, the acceleration changes velocity (either Dec or Inc based on whether it is in same or opposite direction).

So g = 9.8m/s^ 2 means that acceleration of body for every passing second will change by 9.8 m/second.

We know that body will take 2 second to reach the top. At top velocity = 0.

Now comes the gravity. It will increase the velocity of ball such that it becomes 9.8m/s at end of 1st second of downfall.

Now from T=0 to T= 1, the velocity keeps on increasing with every passing fraction of time. so to calculate the distance covered in 1st second we assume the velocity to be constant such that no acceleration is there. This constant velocity is nothing but the average of initial and final velocity which is 0+9.8/2 = 4.9m/s

This means that in 1st second, distance covered by ball is 4.9 m

NOTE- Skip to OR portion if you don't want your brain to be further fried

We can continue this approach for distance covered in every second by taking average of velocity in the beginning and end of that particular second.

At t= 2 second, ball is at same level from which it was thrown. here the velocity is 19.6 m/s (9.8 after 1second and 9.8 further added, thanks to gravity)

So at T = 3, vel is 19.6+9.8= 29.4 Hence avg vel = (19.6+29.4/2) 24.5 m/sec => From T=2 to T =3, distance covered will be 24.5 meters

Similarly for T= 3 to T=4 ,velocity at T=4 =29.4+9.8= 39.2 distance covered will be 34.3m

Hence the total height will be 24.5+34.3= 58.8 meter( since in two second it will reach ground from top of tower)

OR

We know that distance traveled by object in subsiquent seconds is in ratio of 1:3:5:7.......(All hail Gallileo!!!) So in First second distance is 4.9 meter Hence in third and fourth sec combined distance covered will be 5+7= 12 times distance covered in 1 St second which is 12*4.9 = 58.8 m/second