00:01
Hello, here we throw a ball with an initial speed of 25 meters per second at an angle of teta 0, which is 40 degrees above the horizontal line.
00:16
And the wall is at the distance of 22 meters from the point of throw.
00:23
First of all, in question if we have to calculate how far above the release point, the wall will hit the ball.
00:32
The ball will hit the wall.
00:33
So we have to calculate this delta y.
00:36
Let's do this.
00:38
First of all, it will be easier for us to illustrate this system and let's presume that the ball was released from some initial height y0 at the initial speed v0.
00:54
Let's introduce x's y and x and x and the ball hit the wall somewhere here a distance deep first so let us write down the coordinate the dependence of x of ball on as a function of time so that is vx times time which is v0 x cosine teta zero times time because of here v x doesn't depend on gravity because then the only force which is exhorting here the only force which is exerting is gravity and this is downwards so therefore vx is constant now we can calculate time of the flight so time of the flight when x equals to d so that equals to d divided by v0 plus sine theta zero which is 22 meters over 25 meters per second and cosine of 40 degrees that is 1 .5 or sorry 1 .15 or actually let's keep most significant figures 1 .49 seconds.
02:45
Now let's write down y coordinate as a function of time that is y0 plus v0 sign that data 0 t minus gt squared over 2 therefore delta y equals to y at the moment of t at the end of the flight minus y 0 which equals to v0 or that's basically 25 meters per second times sign of 40 degree times 1...
