A cannonball is fired from a cliff 50 meters high with an...

A cannonball is fired from a cliff 50 meters high with an initial velocity of 100 m/s at an angle of 30° above the horizontal. a) How far from the base of the cliff will the cannonball hit the ground? b) What will be the total time of flight? c) What is the maximum height reached by the cannonball above the ground? Assume acceleration due to gravity g = 9.8 m/s².

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Transcript

0:00 Situation.

00:03 An object gets thrown at some speed v naught at some angle theta.

00:10 Okay, it goes up then it comes down and it goes down by some distance of 50 meters.

00:22 Okay.

00:23 Now since it's falling that means that it's negative, which means that delta y is effectively negative 50.

00:33 Okay.

00:35 And also we know that v naught is equal to 30 and theta is equal to 30 degrees, which means that v naught y is equal to v naught sine theta by 30 sine 30 degrees.

00:56 Okay, so if we want to find the time we need to use this kinematic equation, right, and then we rearrange it to get half g t squared minus v naught y t plus delta y equals 0, right? and then that's just 4 .9 t squared minus v naught sine theta t plus delta y equals 0.

01:31 Or 4 .9 t squared minus 30 sine 30 degrees t minus 50 equals 0.

01:45 Okay, and when you put this into wolfram alpha you get t equals negative 2 .01 seconds and 5 .07 seconds.

01:59 This is the one you want because it's not negative.

02:03 Okay, so then for part b we know that it goes up to some max height and then it falls the rest of the way down to ground level.

02:22 So this is y1 and this is y2.

02:26 We also know that here at max height vy equals 0.

02:32 Okay, so then we can use this kinematic equation to find y1.

02:46 Okay, so that's 2g y1 equals v naught y squared...