# The cricketer

1. Oct 8, 2011

### imy786

1. The problem statement, all variables and given/known data

The distance from the Cricketer to the wall is 70 metres across level ground.
The ball reaches its maximum height of 30 metres as it passes directly above the wall

Both the height and the distance to the wall are measured from the point at which the ball is hit by the cricketer.
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1.Calculate the vertical component of the ball's velocity at the point of hit.

2.Calculate the time taken for the ball to reach the highest point.

3.Hence find the horizontal component of the ball’s initial velocity

4.Use the answers to parts (1) and (3) to calculate the magnitude and direction
of the ball’s initial velocity and of its velocity at the highest point.

The cricketer stands back to a distance of 80 metres. When hitting now, the ball only just clears the top of the wall. Calculate the height of the wall.

2. Relevant equations

Will be completing 1 part at a time.

3. The attempt at a solution

1.Calculate the vertical component of the ball's velocity at the point of release.

v^2 = u^2 + 2as

u = ucosθ
v=0
s=70
a=-9.8

v^2 = u^2 + 2as
0= (ucosθ) ^2 - (19.6 x 70)

1372 = (ucosθ) ^2
37.04 = ucosθ

37.0 m/s is the vertical component of balls velocity. (is this correct for part 1).

2. Oct 8, 2011

### dx

I don't think it is. I get 24.5 m/s for the vertical component of the velocity at the point of release.

3. Oct 8, 2011

### imy786

Would be nice if you can tell me were i went wrong from my calculation above

4. Oct 8, 2011

### dx

I don't quite understand what you did there. Maybe you can expand a little on your reasoning. For one thing the distance that would enter into this calculation is 30m, not 70m. If we call the vertical component of the velocity v, then we can write the following equations

vt - (1/2)g(t)(t) = 30

t = v/g

where t is the time it takes to reach the highest point of the trajectory.

5. Oct 8, 2011

### Villyer

You used 70 as your displacement, but that displacement is horizontal. From your givens, the vertical displacement is 30, which when used gives an answer of 24.2 m/s for the vertical component of velocity at point of release.

6. Oct 8, 2011

### imy786

v^2 = u^2 + 2as

u = ucosθ
v=0
s=30 (correction)
a=-9.8

v^2 = u^2 + 2as
0= (ucosθ) ^2 - (19.6 x 30)

588 = (ucosθ) ^2
24.3 m/s = ucosθ

24.3 m/s is the vertical component of balls velocity.

7. Oct 8, 2011

### imy786

2.Calculate the time taken for the ball to reach the highest point.

v=u+at
0=24.3 - 9.8(t)

t = 24.3/9.8 = 2.47 seconds (is this correct for part 2)

8. Oct 8, 2011

### imy786

duplicate post

Last edited: Oct 8, 2011
9. Oct 8, 2011

### dx

Yeah.

10. Oct 8, 2011

### Villyer

If all possible, I would avoid using calculated data to find a further answer to a problem.

In this case, if you use the equation Δy = vo*t + 0.5*a*t2, and look at the problem in the other direction, the time it takes an object to drop 30 meters, the givens become:
Δy = -30
vo = 0 m/s
a = -9.8 m/s2

-30 = 0 * t + 0.5 * -9.8 * t2
-30 = -4.9 * t2
30/4.9 = t2
$\sqrt{30/4.9}$ = t
2.47s = t

So your answer was right, but say for instance you never corrected your answer for part a, you would have used the wrong vo in part b and have gotten time wrong as well.

11. Oct 8, 2011

### imy786

3.Hence find the horizontal component of the ball’s initial velocity

U=U sinθ
a=0
s=30
v=0

s=ut+(1/2)a(t)(t)

v = 15 / (square root of 588 / 9.8)
=6.06 m/s

horizontal initial velocity is of 6.06m/s (is this correct for part 3)

12. Oct 8, 2011

### imy786

duplicate post

13. Oct 8, 2011

### Villyer

vave = Δx/t (the horizontal component of velocity is constant)

Δx = 30m
t ≈ 2.47s (since time is the only common variable in a projectiles problem, it needs to be solved for then used)

vave = 30/2.47
vave ≈ 12.146

I'm unsure where the majority of the numbers you used in your equation came from.

14. Oct 8, 2011

### dx

The time taken to reach the highest point (i.e. t = 2.47 s) is the same as the time taken to reach the wall, so the horizontal component of initial velocity is simply 70m/t = 28 m/s since there is no acceleration in the horizontal direction.

15. Oct 9, 2011

### imy786

3.Hence find the horizontal component of the ball’s initial velocity

U=U sinθ
a=0
s=30
v=0
t=2.47 seconds

s=ut+(1/2)a(t)(t)
30=U sinθ (2.47) + 0

30/2.47 = usinθ =12.146 m/s (so this is the solution for part 3)

16. Oct 9, 2011

### Villyer

That's right now.

17. Oct 9, 2011

### imy786

4.Use the answers to parts (1) and (3) to calculate the magnitude and direction
of the ball’s initial velocity and of its velocity at the highest point.

answer to part (1) and (3)

ucosθ=24.3 m/s
usinθ =12.146 m/s

magnitude=√((24.3)&^2 + (12.146)^2 =

=√(590.49 + 147.53) = 27.16 m/s

tanθ = 12.146/24.3
θ=26.6 degrees is the direction

(is this correct for part 4)

18. Oct 9, 2011

### imy786

duplicate

19. Oct 9, 2011

### Villyer

Actually, looking at what I did earlier, I noticed I used 30 instead of 70 for my change in x.
You did the same, probably from looking at my error. Redo the calculation you did for the horizontal component, but use 70 instead of 30.

20. Oct 10, 2011

### imy786

3.Hence find the horizontal component of the ball’s initial velocity

U=U sinθ
a=0
s=70 (correction)
v=0
t=2.47 seconds

s=ut+(1/2)a(t)(t)
70=U sinθ (2.47) + 0

70/2.47 = usinθ =28.34 m/s (so this is the solution for part 3)