this is an AS physics question about acceleration and velocity.a ball is kicked towards goal posts from a position 20m from and directly in front of the posts.the ball takeds 0.60s from the time it is kicked to pass over the cross-bar,2.5m above the

this is an AS physics question about acceleration and velocity.a ball is kicked towards goal posts from a position 20m from and directly in front of the posts.the ball takeds 0.60s from the time it is kicked to pass over the cross-bar,2.5m above the
this is an AS physics question about acceleration and velocity.
a ball is kicked towards goal posts from a position 20m from and directly in front of the posts.the ball takeds 0.60s from the time it is kicked to pass over the cross-bar,2.5m above the ground .the ball is at its maximum height as it passes over the cross-bar.you may ignore air resistance.
a)calculate the ball's horizontal component of velocity.
b)calculate the vertical component of the velocity of the ball immediately after it is kicked.
c)determine the magnitude of the initial velocity of the ball immediately after it is kicked.
d)determine the angle above the horizontal at which the ball is kicked.

this is an AS physics question about acceleration and velocity.a ball is kicked towards goal posts from a position 20m from and directly in front of the posts.the ball takeds 0.60s from the time it is kicked to pass over the cross-bar,2.5m above the
Hi there.I just got an A* in the entire A level Physics course,so my answer should be trustworthy.:) The reason why I wrote my answer in English was to make it consistent with your question.:P
First of all,the only force exerted on the ball after the kick is the gravitational attraction force.In other words,it experiences free fall.So:
a) Firstly look in the horizontal direction.Assuming that the air resistance is negligible,then there is no horizontal force acting on the ball.Thus the horizontal acceleration is zero (according to Newton's II law).Using v=s/t in the horizontal direction,vh=20/0.6 = 33.3 m/s (vh stands for hotizontal velocity)
b) Now let's look in the vertical direction upwards,the acceleration a of the ball is -9.81m/s^2.(Notice the negative sign,it is there because gravity directs downwards but we are now looking at the upward direction.) Using one of the SUVAT equations s=ut+1/2 at^2 (s=2.5m,t=0.6s,a=-9.81 m/s^2),then we can find out the value for u,which is the initial vertical component of the ball's velocity.
c) Resolving backwards the horizontal and vertical components of the velocity using Pythagora's thereom:V^2= vh^2+u^2.Substitute the values of vh and u which have been worked out previously,you can get the value of V (the magnitude of the initial velocity).
d) Draw a vector triangle for this question would help you a lot.Vertically up is the value of u,in the horizontal direction it is the vector of vh.Complete this triangle.Then you get tan α (the angle above the horizontal) = u / vh.

一个球是踢向球门20米位置,并直接在前面的帖子。球takeds 0.60年代的时间它是踢过去的横杆,2.5米以上的地面。球是在其最大高度经过了横杆。你可以忽略空气阻力。　　一)计算球的水平速度分量。　　b)计算垂直分量的球的速度后就立即被踢。　　c)确定初始速度的大小的球后就立即被踢。　　d)确定角高于水平在球被踢出。...

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一个球是踢向球门20米位置,并直接在前面的帖子。球takeds 0.60年代的时间它是踢过去的横杆,2.5米以上的地面。球是在其最大高度经过了横杆。你可以忽略空气阻力。　　一)计算球的水平速度分量。　　b)计算垂直分量的球的速度后就立即被踢。　　c)确定初始速度的大小的球后就立即被踢。　　d)确定角高于水平在球被踢出。

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