A golf ball is struck at ground level the speed of the golf...

A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in Fig. $4-36$ where $t=0$ at the instant the ball is struck. The scaling on the vertical axis is set by $v_{a}=19 \mathrm{m} / \mathrm{s}$ and $v_{b}=31 \mathrm{m} / \mathrm{s}$ (a) How far does the golf ball travel horizontally before returning to ground level? (b) What is the maximum height above ground level attained by the ball?

Transcript

00:01 This is chapter 4 problem number 38.

00:03 We're given the velocity versus time graph all by golf bowl.

00:09 And the a here is given to us as 19 meters per second.

00:15 And vb is 31 meters per second.

00:18 And in part a, we're asked how far does the wall travel horizontally before returning to ground level? because it was struck from the ground level, right? assume that this is the trajectory of the golf wall and we're given the velocity as a function of time which means at point at time um 2 .5 seconds we're at the top and at the top the velocity that we have is v a and you know that from the project down motion v a has to be equal to the initial x component of the initial velocity if this is initial velocity right and um now, in part a, since we're asked the horizontal displacement, which is, i'm going to call range, range is going to be equal to the x component of velocity times time that it takes from the takeoff until the gulf hole hits the ground again.

01:15 So v .0x is va times t, t being the total time, right? this was the half time when it reaches the top.

01:26 T -print, let's call it, is the total time, okay? we're going to put the t prime here to total time.

01:32 Va is 19 meters per second, time is five seconds.

01:36 So from here, we find it to be 95 meters, the horizontal displacement of the ball.

01:42 Now in part b, we're asked the maximum height, which the bowl reaches.

01:48 So here, this is what we're asked, this h.

01:51 Again, we know that it gets there, the ball gets there in 2 .5 seconds.

01:56 Remember the height equation, since distance traveled in the y direction, is going to be equal to the initial component of the velocity times t minus one -half gt square.

02:08 This delta y means delta minus final that in y initial, pardon me, y -final minus y initial.

02:15 Initial is zero because if the ball takes off from the ground, y -final would be our h.

02:19 We can solve this equation in order to get h...

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