Question

please need help asap

8 A stone is thrown straight up from ground level with an initial velocity 19.6m/s. (Use g = 9.8m/s2). a) How long does it take to reach the maximum height? b) Calculate the maximum height. (4poins)

Answer 1

Given initial vertical velocity, $u_{y}=19.6m/s$

(a) Time to reach the maximum height

At maximum height, vertical velocity becomes zero

Use formula $v=u+at$

$v_{ymax}=u_{y}-gt$

$0m/s=19.6m/s-9.8m/s^{2}*t$

$0=19.6-9.8*t$

$19.6=9.8*t$

$\frac{19.6}{9.8}=t$

ANSWER: ${\color{Red} t=2s}$

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(b) Maximum height

Use formula $s=ut+\frac{1}{2}at^{2}$

$H=u_{y}t-\frac{1}{2}gt^{2}$

$H=19.6m/s*2s-0.5*9.8m/s^{2}*(2s)^{2}$

$H=19.6*2-0.5*9.8*2^{2}$

ANSWER: ${\color{Red}H=19.6m}$

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