# Hello guy's I need help... Can you help me?   A stone is thrown upward with an initial speed of 16 m/s. a.) What will be it's maximum height ? b.) When will it returns to the ground ? c.) Where will it be in 0.8 sec. ? d.) Where will it be in 2.4 sec. ?

## Expert Answers Hello!

Yes we can:)

If we ignore air resistance then a stone will move with the fixed acceleration `g=9.8 m/s^2` (the gravity acceleration). It is directed downwards while the initial speed is directed upwards.

So the speed will be `V(t)=16-g*t` (uniform acceleration) and the height will be

`H(t)=16t-g *(t^2)/(2).`

It is simple to answer all questions using this formula.

a) the maximum for a parabola is reached at `t=-b/(2a)=-(16)/(-2*g/2)=16/g,` the height will be `16*(16)/(g)-(16^2)/(2g)=16^2/(2g) approx 13 (m).`

b) when `H` will be zero for positive `t,` i.e. when `16t-g *(t^2)/(2)=0,` so `t=32/(g) approx 3.3 (s).`

c) `H(0.8)=16*0.8-9.8*(0.8)^2/2 approx 12.8-3.1 approx 9.7 (m).`

d) `H(2.4)=16*2.4-9.8*(2.4)^2/2 approx 38.4-28.2 approx 10.2 (m).`

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