You have :
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v0 = 35 m/s
B = angle of elevation = 30 deg
a suby = - g
v suby0 = ( v0 ) ( sin B )
v suby0 = ( 35 m/s ) ( sin 30 ) = 17.5 m/s
At maximum height :
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v suby = vy* = 0.0
t = t*
y = ymax = y*
Basic kinematics gives :
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a suby = dv suby / dt = -g ------> dt = dv suby /a suby
v suby = dy/dt ------> dt = dy / vsuby
dt = dv suby / a suby = dy / v suby
( v suby ) ( d v suby) = ( a suby ) dy = ( -g ) dy
Now integrate and get :
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( v suby* )^2 - ( v suby0)^2 ( 1/2 ) = ( -g ) ( ymax )
ymax = [ ( v suby0 )^2 - ( v suby* )^2 ] / [ ( 2 ) ( g ) ]
ymax = ( 17.5 m/s )^2 - ( 0.0 m/s )^2 ] / [ ( 2 ) ( 9.807 m/s^2 ) ]
ymax = 15.6 m <------------------
Check the result:
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t* = v suby0 / g = ( 17.5 m/s ) / ( 9.807 m/s^2 ) = 1.784 s
y* - y0 = ( v suby0 ) ( t* ) - ( g/2 ) ( t* )^2
y* = ( 17.5 m/s ) ( 1.784 s ) - ( 9.807 m/s^2 / 2 ) ( 1.784 s )^2
y* = ymax = 31.23 m - 15.61 m = 15.6 m
The ymax values agree.
