An object projected vertically up from the top of the tower took 5s to reach the ground. if the average velocity of the object is 5m/s , find it's average speed.

Question

anonymous · Answer

is the ans is 24.5m/s if it is correct i will explain

anonymous · Answer

it s 5 m/s . is it correct ?

anonymous · Answer

assuming the average velocity is horizontal

9.8*5 = 49m/s
since constant acceleration average vertical speed =24.5 m/s.


now taking vertical and horizontal components the average speed = root(5^2+24.5^2)=25.0 m/s (1d.p)

anonymous · Answer

(the last bit was done using pythagorus, i realised i didn't write enough of the steps down and i didn't want you to get confused. shoiuld have written before that s^2=5"+24.5^2 where s=speed, 5=average horizontal speed and 24.5=average vertical speed. sorry if that felt patronising it was not intended its just i lose alot of peole when i explain my methods. even my teacher who's a doctor of physocs and applied maths!)

anonymous · Answer

It takes 3 seconds to go up and 2 to come down (roughly).
average speed is +12.78 ms  assuming + is up.
29.52 m/s V initial and -19.25 V final.  Plot velocity vs time v= vo+gt
straight line. integral of V from 0-5 s must equal avg V time (delta t)