(a)
Use kinematic equation,
v = u + at
where v is final speed ,u is initial speed ,a = -g and t = time
0 = 22 m/s - 9..8 t
Solve for t
t = 2.24 s
This is time of ascent. Total time in air is twice the time ascent.
2t = 4.48 s
Round off into 2 significant digits
2 t = 4.5 s
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(b)
From kinematic equation
v^2 = u^2 +2aS
0 = ( 22 m/s ) ( 22 m/s ) -2 (9.8 ) S
Solve for S
S = 24.44 m
Round off into 2 significant digits
S = 24 m
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(c)
From kinematic equation
S = ut +1/2at2
24.44 m = 22 t - 1/2 *9.8 *t2
4.9 t2 -22 t +24.44 =0
Solve quadratic equation for t
t = 2.01 s ( first time ) and 2.47 ( second time )
Round off into 2 significant digits
t = 2.0 s ( first time ) and 2.5 s ( second time )
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