Ask Me Help Desk - Kinematics-acceleration,free fall,etc.

1. The maximum deceleration of a car on a pavement is about 10m/s^2. What would the police conclude about a driver who left skid marks 80 m long?

2. A raindrop falls from a cloud 100 m above the ground. Neglecting air resistance, what is the speed of the raindrop when it hits the ground?

3. A golfer hits his tee shot along a flat fairway at 40 degrees to the horizontal with an initial speed of 50 m/s. What is the range of the ball? What is the maximum height to which the ball rises?

4. An archer tries to hit a target that is 20 m away from him. He can release the arrow at 25 m/s. Neglecting air resistance, estimate the angle at which the archer should aim to compensate for the fall of the arrow due to gravity.

5. Trailing by two points, and with only 2.0 seconds left in the last quarter of a basketball game, a player makes a jump shot at an angle of 70 degrees with the horizontal giving the ball a velocity of 10 m/s. The ball is released at the height of the basket and it's a score! Did the player tie the game or put the team ahead?

Ok, but you have to try, I've already given you the 'how to work out'.

1. Using $v^2 = u^2 + 2as$ v being the final velocity, u the initial velocity, a the acceleration and s the displacement;

$(0)^2 = u^2 + 2(-10)(80)$

$1600= u^2$

$40= u$

Therefore the initial velocity of the car is 40 m/s

2. Again using $v^2 = u^2 + 2as$ ,

$v^2 = (0)^2 + 2(9.81)(100)$

$v^2 = 1962$

$v = 44.3$

Final speed = 44.3 m/s

3. Vertical displacement = 50 sin40 = 32.1 m/s
Horizontal Displacement = 50 cos40 = 38.3 m/s

Using $v^2 = u^2 + 2as$ ,

$(0)^2 = (32.1)^2 + 2(-9.81)s$

$s =52.5 m$

Max height = 52.5 m

Using $s = ut + \frac12 at^2$
Time for ball to come back down:

$(0) = (32.1)t + \frac12 (-9.81)t^2$

$32.1 -4.905t = 0$

$t = 6.54 s$

Horizontal distance covered in 6.54 s = 38.3 x 6.54 = 446.7 m

4. I remembe having done a similar question like this before. I just checked it, and a formula was given to solve the problem, a formula you can derive.

Ok, it's $d = \frac{v^2sin(2\theta)}{g}$

Plug in your given values; $(20) = \frac{(25)^2sin(2\theta)}{9.81}$

And you should obtain 9.15 degrees with the horizontal.

5. Horizontal velocity = 10 cos70 = 3.42 m/s
Vertical velocity = 10 sin70 = 9.40 m/s

Time of flight (consider vertical only);

$0 = (9.40)t + \frac12 (-9.81)t^2$

$0 = 9.40 - 4.905t$

$t = 1.92$

Therefore range = 3.42 x 1.92 = 6.55 m (keeping all the digits from the start, that is 3.4202... and 1.9157... )

Do you know about the rules of basketball concerning marking? I guess you do, and it's obvious that you know whether it's a tie or the player put the team ahead.

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Originally Posted by Unknown008

Ok, but you have to try, I've already given you the 'how to work out'.

1. Using $v^2 = u^2 + 2as$ v being the final velocity, u the initial velocity, a the acceleration and s the displacement;

$(0)^2 = u^2 + 2(-10)(80)$

$1600= u^2$

$40= u$

Therefore the initial velocity of the car is 40 m/s

2. Again using $v^2 = u^2 + 2as$ ,

$v^2 = (0)^2 + 2(9.81)(100)$

$v^2 = 1962$

$v = 44.3$

Final speed = 44.3 m/s

3. Vertical displacement = 50 sin40 = 32.1 m/s
Horizontal Displacement = 50 cos40 = 38.3 m/s

Using $v^2 = u^2 + 2as$ ,

$(0)^2 = (32.1)^2 + 2(-9.81)s$

$s =52.5 m$

Max height = 52.5 m

Using $s = ut + \frac12 at^2$
Time for ball to come back down:

$(0) = (32.1)t + \frac12 (-9.81)t^2$

$32.1 -4.905t = 0$

$t = 6.54 s$

Horizontal distance covered in 6.54 s = 38.3 x 6.54 = 446.7 m

4. I remembe having done a similar question like this before. I just checked it, and a formula was given to solve the problem, a formula you can derive.

Ok, it's $d = \frac{v^2sin(2\theta)}{g}$

Plug in your given values; $(20) = \frac{(25)^2sin(2\theta)}{9.81}$

And you should obtain 9.15 degrees with the horizontal.

5. Horizontal velocity = 10 cos70 = 3.42 m/s
Vertical velocity = 10 sin70 = 9.40 m/s

Time of flight (consider vertical only);

$0 = (9.40)t + \frac12 (-9.81)t^2$

$0 = 9.40 - 4.905t$

$t = 1.92$

Therefore range = 3.42 x 1.92 = 6.55 m (keeping all the digits from the start, that is 3.4202..... and 1.9157....)

Do you know about the rules of basketball concerning marking? I guess you do, and it's obvious that you know whether it's a tie or the player put the team ahead.

I thought in number three the formulas to be used are the following:

R = [u^2 sin 2 (angle measurement)]/g
maximum height or dy = (u sin angle)^2/2g
t = [u sin (angle)]/g

In #3, why is the acceleration due to gravity becomes -9.81 m/s? Also you already found the value of s, why is it it became 0 in finding the time?

Horizontal distance covered in 6.54 s = 38.3 x 6.54 = 446.7 m ?

Or we can use t = [2(32.14 m/s)]/ 9.81 m/s^2

Fyi, I had already answered the first three qs, well, I was wrong in #3 cos I used the wrong formulas...

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Horizontal distance covered in 6.54 s = 38.3 x 6.54 = 446.7 m ?

Oh gosh! I don't know what happened for number 3. I'm using an old calculator and maybe it picked another value. Anyway, I did that again and I got 250 m (3 sf)

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I thought in number three the formulas to be used are the following:

R = [u^2 sin 2 (angle measurement)]/g
maximum height or dy = (u sin angle)^2/2g
t = [u sin (angle)]/g

For your next questions, in Mauritius, all the teachers tell us it is not necessary to learn the formulae you just said. That's because our examiners will never ask us to give that or that formula and these will not be considered in "working" marking.

However, if you passed through the different steps, finding the time, height, vertical and horizontal displacements, you will have those marks.

Moreover, when you take the answer you got from a previous part, that's risky in the case that you did a mistake.

Say that number three. I did by mistake the maximum height before the range. If I did the range first, I would know the time of flight of the ball. Half of which would give the time when the ball will be at maximum height, which makes the next part easier. Consider a situation where I did a mistake in finding the time, using that 'wrong' time, I would get the wrong answer for the next part.

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In #3, why is the acceleration due to gravity becomes -9.81 m/s? Also you already found the value of s, why is it it became 0 in finding the time?

The 's' here means the vertical displacement. In the first instant, at maximum height, s is maximum, of course. Now, to find t, I must have displacement zero, that is the time for the ball to go up, then down. That is the total time. I therefore have to consider the displacement to be zero.

That displacement is zero at the start, and when it hits the ground (considering the vertical only.

Quote:

Or we can use t = [2(32.14 m/s)]/ 9.81 m/s^2

You can use that only when the initial velocity is equal to the negative final velocity. That's because you get it from:

$v = u + at$

If u = -v, then,

$2v = at$

Quote:

Fyi, I had already answered the first three qs, well, I was wrong in #3 cos I used the wrong formulas...

Yes, I haven't checked my mail yet that day... :o

Hope it helped! :)

Oh, and your other question:

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Is 9 degrees 1 minute and 49.76 seconds also correct?

9.1478137... is not 9 degrees 1 minute and 49.76 seconds but 9 degrees 8 minutes and 52.13 seconds from my calculator...