Question

A 2-kg object is fired from a cannon at a 30º angle wrt to the horizontal. the initial speed is 35 m/s. how high will the object travel?

Answer 1

     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.