00:01
Hi there, here we have to solve the following problem.
00:03
A golf ball is hit at a speed of v0, that is 35 meters per second, at an angle of 50 degree above the horizontal.
00:19
So the target is 80 meters from the launch point, and first we have to calculate the height which the ball hits during the strike.
00:37
Let's introduce the axis.
00:46
So, and we have to find h maximum.
00:50
So to do this, we need to first calculate the component of the initial velocity.
01:00
So v0y equals to v0 times sine alpha.
01:10
Y component of the velocity changes according to the following equation, that is v0y minus gt, because of the component of the velocity changes according to the following equation, because gravity is acting downwards.
01:30
Then, time which corresponds to the maximum height is a shift when a vertical component of the speed is zero.
01:43
So therefore, this time equals to v0 sine alpha over g.
01:56
On the other hand, the maximum height equals to v0 y squared divided by 2g which is v0.
02:10
Sine alpha that is squared over 2g.
02:38
Let's calculate this number.
03:01
That is 33.
03:09
36 .7 meters.
03:19
So we answered question a.
03:21
Now let's move on to question b, where we have to calculate speed.
03:26
Of the ball when it hits when it hits the wall.
03:33
So first of all, x component of the speed is constant because there is no force acting in the y direction and that equals to 35 .0 meters per second times cosine of 50 degree.
04:10
That is 19 .3 meters per second.
04:17
So now we have to find x, sorry, we have to find vy...