00:01
For this problem, you're considering a ball that is kicked from ground level at some initial velocity.
00:10
You're told that it clears a fence at the edge of the field and you want to find the time it took for the ball to reach the fence, the height of the fence, and the distance beyond the fence where the ball lands.
00:22
So we'll start by drawing a picture here of what's happening.
00:26
So if this is the ground, the ball is initially kicked from the ground.
00:32
At some angle theta knot and we're told that that initial angle is 42 degrees above the horizontal.
00:45
It has a speed b knot that we're told is 14 meters per second.
00:55
And then sometime later it's going to start to fall.
01:01
Eventually it hits the ground, but before it hits the ground, it is going to just clear a fence.
01:12
So there's my fence.
01:14
The fence is 18 meters from where the ball was kicked.
01:18
So we know that this distance here, which i'm going to label as delta x, is 18 meters.
01:29
And that's all the information that is given.
01:32
So for part a, you want to know the time it took the ball to reach the fence.
01:37
We're going to just use the kinematic equations and specifically the kinematic equation for horizontal motion, which tells us that the distance traveled in the horizontal direction is equal to the initial velocity in that direction times time.
01:59
So we can, of course, solve for t and get that time then must be equal to delta x over the initial velocity in the x direction, which can be, be written as v -0 cosine of theta -not.
02:15
And so we just go ahead and plug in the values that were given in the problem, and you would find that it takes about 1 .73 seconds for the ball to reach the fence.
02:28
For part b, we want to know what is the height of the fence.
02:33
And so the height of the fence is going to be the same as the vertical displacement of the ball.
02:39
Once it's at that location.
02:43
And so again, we'll use the kinematic equations, but now in the vertical direction for a projectile.
02:50
And specifically that the vertical displacement, delta y, is equal to initial velocity in the y direction times t minus one half gt squared.
03:06
And again, we'll make the substitution, that the initial velocity in the y direction is going to be the initial velocity times sign of the initial angle and we'll use that time of 1 .73 seconds that we found in part a to plug in and get that delta y should be about 1 .54 meters.
03:38
Lastly for part c we want to know that the distance beyond the fence where the ball lands...