You tip your cannon to an angle of 500 above the ground and...

You tip your cannon to an angle of 50.0° above the ground and set it on a hill 20.0 m above the valley. It is still shot with an initial velocity of 40 m/s. At what height above the valley floor will the cannonball be after 2.20 s? At what height above the valley floor will the cannonball be after 7.00 s? Find the maximum height. Find the time it takes to reach the groun How far downrange will the cannonball land?

Transcript

00:01 The cannonball is fired at the castle.

00:02 The cannon emerges from the cannon at the height of 7 meters above the ground at the angle of 37 degrees to the horizontal at the launch speed of 40 meters per second.

00:18 So the y component of this velocity is 40, sign 37.

00:22 The x component is 40 cost 37.

00:27 We want to find first of all the maximum height above the ground reached by the cannonball.

00:32 So here we have the initial velocity in the y direction.

00:36 We have the acceleration due to gravity going downwards at the final velocity at the maximum height so we're going to use the equation v squared is equal to u squared plus 2 as and this is in the y direction so the final velocity in the wide direction that's v so that would be 0 minus is equal to the inch of loss in the wide direction that's 40 sign 37 square that and we'll take an opus positive so this is positive and g will be negative there so this is plus two times the acceleration that's g minus 9 .8 to 1 times the displacement which is going to be the height just called a h so h is equal to 29 .54 meters so now for the maximum height about the ground and to add it to the 5 7 meters which you will which was originally which it was originally above the ground at so we're going to have 38 .5 4 it does a alpha b alpha horizontally from the cannon ball does the from the cannon ball land so we're going to find the time of flight by using the equation s equal to u t plus half a t squared so we have s is going to be the displacement which is 7 meters downwards so down is negative that minus 7 and u is our initial losty we're still learning with wide direction here so that is 40 sign 37 square that sir no squared multiplied by the time which is unknown plus half times our acceleration minus 9 8 1 c squared so i'm solving for t here.

02:56 5 .1 .88 seconds.

03:01 Now we're taking this into the x direction with velocity is constant...

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