A a long jumper leaves the ground at 45circ above the...

(a) A long jumper leaves the ground at 45$^\circ$ above the horizontal and lands 8.0 m away. What is her "takeoff" speed $v_0$? ($b$) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m vertically below. If she long jumps from the edge of the left bank at 45$^\circ$ with the speed calculated in ($a$), how long, or short, of the opposite bank will she land (Fig. 3-50)?

Transcript

00:01 All right, someone is jumping 45 degrees.

00:03 By the way, the sign of 45 degrees equals the cosine of 45 degrees, which is going to be the square root of 2 over 2.

00:11 Lands 8 meters away, what's the takeoff speed? well, the distance to the peak would be half of 8 meters.

00:26 That would be 4 meters.

00:27 That would be the x distance to the peak.

00:29 So maybe i should call this x.

00:32 To the peak.

00:36 The time to the peak would therefore be the horizontal speed.

00:48 No, velocity equals distance over time.

00:53 So time is going to equal distance over velocity.

00:58 So the time is going to equal distance over velocity.

01:01 So the time is going to be x to the peak over the horizontal component of the velocity, which would be square root of 2 over 2 times v initial.

01:22 So now, 8.

01:41 Now, i also know that the velocity in the y direction at the peak is going to equal 0, which is going to equal, the y direction velocity, which is the square root of 2 over 2 v0.

02:02 So v equals v initial minus g t to the peak.

02:10 So t to the peak is going to be square root of 2 v0 over 2g.

02:20 So i can now put that up here.

02:23 Square root of 2 v0 over 2.

02:27 Over to g.

02:31 Now, i can multiply by v0 on both sides, and the square root of twos cancel out, by the way.

02:43 Square to 2 over 2, divided by square root of 2 over 2, those will cancel out.

02:48 And so v0 squared is going to be x peak times g.

02:54 So v0 is going to be the square root of x peak.

02:58 Peak times g um so it's going to be the square root of 4g putting that in the calculator v equals a square root of 4g 6 .3 meters per second okay it's 10 so now she needs to jump 10 meters and a half meters below.

04:07 How long or short will she land? okay, so she's jumping from up here.

04:15 She needs to get over here.

04:22 We know v0.

04:24 So, y, final, needs to be negative 2 .5.

04:38 Equals y initial, which is 0, plus v initial.

04:44 In the y direction, which would be this v0 times a square root of 2 over 2 times t minus one half gt squared to go the whole way.

05:03 X final, which we're going to try to figure out, it's going to be x initial, which is zero, plus vx initial, which would be v0 squared a 2 over 2 times t...

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