Projectile Motion Q22 Fitz3U (1 Viewer)

[azureus88]

"A batsman hits a cricket ball 'off his toes' towards a fieldsman who is 65m away. They ball reaches a maximum height of 4.9m and the horizontal component of its velocity is 28m/s. Find the constant speed with which the fieldsman must run forward starting at the instant the ball is hit, in order to catch the ball at a height of 1.3m above the ground."

I tried out this question but cant seem to work out what i did wrong. Can someone help me spot the flaw in my working. thanks!

Let eqation of path of ball be

x^2 = 4a(y-4.9)

when x=32.5, y=0
1056.25 = 4a(-4.9)
a= -21125/392

Subbing back into equation, x^2 = -21125/98(y-4.9)

when y=1.3, x=(27 and 6/7)

time taken for ball = time taken for fieldsman to run

[32.5+(27 and 6/7)]/28 = [32.5-(27 and 6/7)]/v where v is required speed

Rearranging, v = 2.15m/s (2dp)

which is incorrect answers 7m/s, what did i do wrong??

 

[lyounamu]

"A batsman hits a cricket ball 'off his toes' towards a fieldsman who is 65m away. They ball reaches a maximum height of 4.9m and the horizontal component of its velocity is 28m/s. Find the constant speed with which the fieldsman must run forward starting at the instant the ball is hit, in order to catch the ball at a height of 1.3m above the ground."

I tried out this question but cant seem to work out what i did wrong. Can someone help me spot the flaw in my working. thanks!

Let eqation of path of ball be

x^2 = 4a(y-4.9)

when x=32.5, y=0
1056.25 = 4a(-4.9)
a= -21125/392

Subbing back into equation, x^2 = -21125/98(y-4.9)

when y=1.3, x=(27 and 6/7)

time taken for ball = time taken for fieldsman to run

[32.5+(27 and 6/7)]/28 = [32.5-(27 and 6/7)]/v where v is required speed

Rearranging, v = 2.15m/s (2dp)

which is incorrect answers 7m/s, what did i do wrong??

I will have a shot at this.

Max height = 4.9

Maximum height = (y^2 . sin^2(@))/2g = 4.9
y^2 . sin^2(@) = 96.04
ysin@ = SQRT(96.04) = 9.8... = 9.8 m/s

Now, y = -1/2 gt^2 + vsin@t t
= -4/9 t^2 + 9.8t

Ball arrives at a height of 1.3 (i.e. y=1.3):
1.3 = -4/9 t^2 + 9.8 t
Rearrange:

4/9 t^2 - 9.8t + 1.3 = 0
Using the quadratic formula (cannot be bothered to write the whole thing out)
t = 0.14..s or 1.86...

Sub both time then you will see that t = 0.14 is unrealistic.

So when t=1.86, x = 52
which means that the cricket player must run 13 m in 1.86.

So S = d/t = 13/1.86 = 6.98924... = 7

 

[azureus88]

thanks for the solution, but wat was wrong with the catesian method?

 

[davidbarnes]

I wouldn't use the formulas at all. The cartesian method is the way to go and the way they'll ask you to do a question in the exam in general.

 

[lyounamu]

I wouldn't use the formulas at all. The cartesian method is the way to go and the way they'll ask you to do a question in the exam in general.

I didn't use the formula, I actually derived them myself but I just couldn't be bothred to show all the working...

 

[lyounamu]

thanks for the solution, but wat was wrong with the catesian method?

Did you do it properly though?

 

[azureus88]

oh...no wonder. i get it now. thanks for all the help, greatly appreciated.

 

[QuLiT]

Does anybody else use their physics knowledge to 'check' projectile motion answers?

lol i use maths to check physics answers xD