antiderivatives (motion): With what initial velocity must an

[wonky-faint]

With what initial velocity must an object be thrown upward (from a height of 2 meters) to reach a maximum height of 200 meters?

neglect air resistance; use a(t)=-9.8m/s for acceleration due to gravity

this is what i did:

-4.9t^2+v(0)t+s(0)=s

-4.9(0)^2+v(0)t+2=200
v(0)t=198

then how do i find v(o) if i dont know time?
is what i did so far even right?

(i wasnt sure to use 0 for acceleration as it would be 0 for when the object is at maximum, or to use -9.8)

 

[skeeter]

s = s_o + v_ot - 4.9t²

it's going to take a bit more than "0" seconds to reach a height of 200 meters, and the acceleration is a constant -9.8 m/s².

find velocity by taking the derivative of the position function ...

v = v_o - 9.8t

at its maximum height, the object has a velocity of 0 ...

0 = v_o - 9.8t

t = v_o/9.8

now, substitite (v_o/9.8) for t in the position equation ...

200 = 2 + v_o(v_o/9.8) - 4.9(v_o/9.8)²

solve for v_o

 

[wonky-faint]

oh thankyou!! you made it look so easy....

i think when i plugged in 0 for t, i meant to plug it in for a in the equation s=1/2at^2+v(0)t+s(0)

where i plugged in -9.8 for a, and got -4.9. i wasnt sure that instead there i should plug 0 for a as acceleration at maximum is 0, so that part would become 0.

i mixed the two up there. sorry.

but your way helped so thanks![/code]