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October 1, 2014

October 1, 2014

Posted by **Elle** on Thursday, October 20, 2011 at 4:17am.

The velocity of the ball as it leaves the kicker’s foot is 16 m/s at angle of 46◦ above the horizontal. The top of the fence is 4 m high. The ball hits nothing while in ﬂight and air resistance is negligible. The acceleration due to gravity is 9.8 m/s^2.

Determine the time it takes for the ball to reach the plane of the fence.

Answer in units of seconds.

How far above the top of fence will the ball pass? Consider the diameter of the ball to be negligible

What is the vertical component of the velocity when the ball reaches the plane of the fence? Answer in units of m/s

Please help, this is college physics!!! thank you!

- Physics.. please help!!!!! -
**drwls**, Thursday, October 20, 2011 at 6:03amThe horizontal velocity component, which does not change, is

Vx = Vo cos46 = 11.11 m/s. Divide the distance to the fence by Vx to obtain the flight time to the plane of the fence.

Use that time (t) and the equation for height above the ground (given below), to find out how far whether, or by how much, the ball clears the fence.

y = (16 sin46)t - (g/2)*t^2

For the vertical component of V,

Vy = 10 sin46 - gt

- Physics.. please help!!!!! -
**Amanda**, Monday, October 8, 2012 at 6:35pmThat is not right drwls

- Physics.. please help!!!!! -
**Drew**, Tuesday, January 29, 2013 at 8:52pmThe third question is wrong...

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