A ball is thrown vertically with a speed of 250 ms from a...

A ball is thrown vertically with a speed of 25.0 m/s from a height of 2.0 m above the ground a) how high does the ball rise b) how long does it take to reach its highest point c) how long does it take to hit the ground after it reaches its highest point d) what is the ball's velocity when it returns to the level from which it was thrown

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00:01 To solve this question, first let's write whatever it is given.

00:07 Initial speed which is equal to 10 m per second.

00:12 Speed at highest point which is v, the final point is 0 m per second because the ball is dropping down.

00:21 The height reach h is h and the time reach the highest point which is the time of ascent and velocity at the same point from which the ball was projected is same as v.

00:42 So firstly to solve this a part, we will solve this by using the third equation which states that v2 is equal to u2 plus 2 times a times s.

01:02 Since motion is against gravity, then acceleration is minus g because it is against gravity.

01:09 So a resistance will be there.

01:12 0 is equal to 10 .2 plus 2 times negative 9 .8 times h because we need to find the height.

01:25 So height is equal to 100 by 19 .6 which is equal to 5 .10 meters.

01:38 So now to solve the b part.

01:41 We need to use the first equation which is equal to v is equal to u plus at.

01:48 At means a times t, acceleration times time.

01:54 0 is equal to 10 plus, since the acceleration is against gravity, so it is negative.

02:02 So acceleration because gravity, acceleration is equal to negative gravity which is negative 9 .8 times t.

02:11 T is equal to 10 by 9 .8 which is equal to 1 .02 seconds.

02:19 So now to solve the c part...

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