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
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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 !!☺☺☺☺☺
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 !!☺☺☺☺☺
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