A ball is thrown directly downward with an initial speed of 8.00 m/s, from a height of 30.0 m. After what time interval does it strike the ground?

Question

A ball is thrown directly downward with an initial speed of 8.00 m/s, from a height of 30.0 m.  After what time interval does it strike the ground?

mathmate · Answer

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hint: use the kinematics equation
S=ut+(1/2)at^2
S=-30 m  distance travelled (negative downwards)
u=-8 m/s   initial velocity, (negative downwards)
t=time in seconds (to be solved for)
a=-9.81 m/s^2   acceleration due to gravity, negative downwards)
Solve for t as the time required, reject negative root (for time).

kofi · Answer

alright thank you

mathmate · Answer

you're welcome!  :)

ravneet · Answer

Use  V^2=U^+2AS to find final velocity and than apply V=U+AT where
V=final velocity
U=Initial Velocity(-8)
A=Acc. due to gravity(-10 approx)
S=displacement(-30)
T=time of motion [ all are negative as I have taken downward direction as negative and Upward direction as Positive . you can change their sign according to your convience ]
NOW FIND time SO NO NEED TO SOLVE QUADRATIC EQUATION