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a projectile is launched upward from ground level with an initial speed of 98m/s.How high will it go?When will it return to the ground​

Sagot :

Answer:

489.5 m, 20s

Step-by-step explanation:

s= vot - 1/2 gt2 1st equation

vf=vo-gt 2nd equation

If it reach the peak its velocity is 0 (vf),

Know that g (gravity) is 9.81m/s2

Substituting it to 2nd eq

0=98-9.81t

t=98/9.81

t=10

Substituting to 1st equation

s=98(10)-1/2 (9.81)(10)^2

s or the height it reach =489.5m

The time to go up is 10s and same 10s to go down, so 10+10 =20s

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