an object is thrown straight up from the top of a tall building with a speed of 10.5 m/s. how high above or how low below the projection point is the object after 2.5s?

Question

anonymous · Answer

acceleration due to gravity=9.8m/s2
initial velocity=-10.5m/s (the negative sign means that the object is moving against gravity)
and time taken  is 2.5m/s
With this information, can you figure out which equation to use to find the distance?

sumeet · Answer

something is wrong with your data. time taken can't be m/s...

anonymous · Answer

oh, yeah its 2.5s
unintentional mistake

sumeet · Answer

First use $$v = u + at$$
where v= velocity after time t
          t = time lapsed
          u =initial velocity
object is thrown upwards to g=-9.81
u=10.5 m/s

Calculate v, i.e. velocity after 2.5 s
Then use
$$v^2 = u^2 + 2as$$
a=-9.81
u=10.5m/s
v= as found out using above equation
s=displacement in m

anonymous · Answer

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