Asked by physics
                A place kicker must kick a football from a point 36 m (about 39 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 24 m/s at an angle of 50 to the horizontal.
The acceleration of gravity is 9.8 m/s2.
To determine if the ball approaches the crossbar while still rising or while falling, what is its vertical velocity at the crossbar?
Answer in units of m/s
            
        The acceleration of gravity is 9.8 m/s2.
To determine if the ball approaches the crossbar while still rising or while falling, what is its vertical velocity at the crossbar?
Answer in units of m/s
Answers
                    Answered by
            Henry
            
    Vf^2 = Vo^2 + 2gd,
Vf^2=(18.39)^2 + (-19.6)*3.05 278.41,
Vf = 16.69m/s at the crossbar.
 
    
Vf^2=(18.39)^2 + (-19.6)*3.05 278.41,
Vf = 16.69m/s at the crossbar.
                    Answered by
            Henry
            
    Disregard above solution.
Vo = 24m/s@50deg.
Xo = hor. = 24cos50 = 15.43m/s.
Yo = ver. - 24sin50 = 18.39m/s.
t(up) = (Yf - Yo) / g,
t(up) = (0 - 18.39) / -9.8 = 1.88s.
hmax = (Yf^2 - Yo^2) / 2g,
hmax = (0 - (18.39)^2) / -19.6=17.25m.
Yf^2 = Yo^2 + 2g*(17.25-3.05),
Yf^2 = 0 + 19.6*14.2 = 278.32,
Yf = 16.69 @ the crossbar.
    
Vo = 24m/s@50deg.
Xo = hor. = 24cos50 = 15.43m/s.
Yo = ver. - 24sin50 = 18.39m/s.
t(up) = (Yf - Yo) / g,
t(up) = (0 - 18.39) / -9.8 = 1.88s.
hmax = (Yf^2 - Yo^2) / 2g,
hmax = (0 - (18.39)^2) / -19.6=17.25m.
Yf^2 = Yo^2 + 2g*(17.25-3.05),
Yf^2 = 0 + 19.6*14.2 = 278.32,
Yf = 16.69 @ the crossbar.
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