Question

A point object is thrown vertically upwards at such a speed that it returns to the thrower after 6 seconds with what speed was it thrown up and how high did it rise

Answer 1

hii mate!!


time in air when thrown vertically upward is given by

t= 2*u/g  [u is initial velocity and g is accln due to gravity]

Now, given t = 6s

  g=9.8 m/s2

6=2*u/9.8

u= 6*9.8/2

u= 3 * 9.8=29.4 m/s

Now, u = 29.4 m/s

  v (final vel. at highest point) = 0 m/s

  g= -9.8 m/s2 (as its opposite to dirn of vel.)

Max. height = (v2 - u2)/ 2g

  = [0 - (29.4)2]/ 2* -9.8

  = -864.36 / -19.6

  = 44.1 m

hope it helps !!☺☺☺☺☺