Asked by geof
a place kicker kick a ball 36m from a goal .which is 3.05 m high.it leaves the ground 20 m/s at an angle 53 to the horizontal.how much the ball clear or fall short of clearing the bar
Answers
Answered by
That's Mr. Idontknowhwhatimdoing to you sir!
Use d = d0 + v0t + (1/2)at^2
In the x direction:
x = x0 + v0t + (1/2)at^2
solve for t
Plug t into:
y = y0 + v0t + (1/2)at^2
y - 3.05 = The distance by which the ball will clear, or fall short of, the bar.
In the x direction:
x = x0 + v0t + (1/2)at^2
solve for t
Plug t into:
y = y0 + v0t + (1/2)at^2
y - 3.05 = The distance by which the ball will clear, or fall short of, the bar.
Answered by
That's Mr. Idontknowhwhatimdoing to you sir!
Velocity in x direction = v0cos(53)
Velocity in y direction = v0sin(53)
Velocity in y direction = v0sin(53)
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