00:01
So let's denote the initial horizontal position xnat of the ball as 0 meter.
00:06
It was stated that the ball leaves truly hand at the height of 1 .5 meters above the ground.
00:11
So this is the initial vertical position of the ball.
00:14
So 1 .5 meter if we denote the ground as the zero vertical position.
00:22
It was stated that the initial velocity of the ball is 11 meter per second and the launching angle is 41 degrees.
00:34
Now question one asks what is the horizontal component of the velocity of the ball when sarah touches the ball and it was stated at a height of 1 .5 meters above the ground.
00:53
To solve for the horizontal component of velocity we will use the climatic equation for horizontal component of velocity which is this and since the ball is in projected motion its motion is due to gravity alone.
01:15
As such, it's the horizontal component of the acceleration is zero, and the equation becomes like this.
01:24
And for a projectile motion, the horizontal component of initial velocities equals to v -nought, cosine theta -nought, like trigonometry.
01:34
Then equation becomes like this.
01:38
In solving, we have 11 meter per second multiplied by cosine 41 degrees is equals to 8 .3.
01:51
Meter per second.
01:53
So this is the horizontal component of the velocity of the wall where sarah catches.
02:01
For question two, what is the vertical component of the velocity? since we don't know that time yet, this is the equation we will use since we know the initial and final vertical positions of the world.
02:32
Then certain the initial position, the final so we have for the final, it's 1 .5 meter and the initial is also 1 .5 meter.
02:43
And note that for the initial component of the velocity, vertical component, is equal to v0, sine, theta, so we have this.
03:05
Then this will become opposed to zero.
03:09
And if we divide both sides by t, we could have the t in both sides of rotation, and multiply the both sides by two, it could also add the two in the right side equation.
03:19
So our equation becomes like this.
03:31
Soving we have negative 11 meter per second by sign 4 to 1 degrees equals to negative 7 .22 meter per second.
03:40
So this is the vertical component of the velocity when sarah catches the ball.
03:46
Number three, what is the time of flat? it's the ball before it was catch.
03:55
So using again the equation for vertical position.
04:00
When we have this, which we know equals to this.
04:20
And since we know the difference of the final and initial position of table is zero, then simplifying our equation become like this.
04:35
And we can see that t is equals to zero and t is equals to b0 .1 over g.
04:46
Since we can equate this to zero and solve for t.
05:21
Then time of flight is this one since take cause serious any shot.
05:29
So we have to be not signed 10a over g...