Projectile Motion Q (1 Viewer)

[learninstudent]

This is from cambridge 3u ex 3g q 12

Glenn the fast bowler runs in to bowl and releases the ball 2.4m above ground with speed 144km/h at angle of 7degrees(*) below the horizontal. Take origin to be point where the ball is released and g=10m/s^2

a) show coordinates of ball t seconds after ts release given by:
x= 40tcos7* and y=2.4-40tsin7* -5t^2
b) how long will it be before ball hits the pitch
c) calculate angle at which the ball will hit the pitch
d) the batsman is standing 19m from point of release. If ball lands more than 5m in front of him, it wil be classified as "short pitched" delivery. Is this particular delivery short pitched?


also.....
q14 - A cricketer hits ball from ground elvel with speed 20m/s and angle of elevation a.. it flies towards a high wall 20m away on level ground. Take the origin at the point where the ball was hit and g=10m/s^2
a) show that the ball hits the wall when h= 20 tan a - 5sec^2 a
c) show max value of h occurs when tan a =2
d) find max height
e) find speed and angle at which the ball hits the wall

thanks in advance =p !

 

[random-1006]

This is from cambridge 3u ex 3g q 12

Glenn the fast bowler runs in to bowl and releases the ball 2.4m above ground with speed 144km/h at angle of 7degrees(*) below the horizontal. Take origin to be point where the ball is released and g=10m/s^2

a) show coordinates of ball t seconds after ts release given by:
x= 40tcos7* and y=2.4-40tsin7* -5t^2
b) how long will it be before ball hits the pitch
c) calculate angle at which the ball will hit the pitch
d) the batsman is standing 19m from point of release. If ball lands more than 5m in front of him, it wil be classified as "short pitched" delivery. Is this particular delivery short pitched?


also.....
q14 - A cricketer hits ball from ground elvel with speed 20m/s and angle of elevation a.. it flies towards a high wall 20m away on level ground. Take the origin at the point where the ball was hit and g=10m/s^2
a) show that the ball hits the wall when h= 20 tan a - 5sec^2 a
c) show max value of h occurs when tan a =2
d) find max height
e) find speed and angle at which the ball hits the wall

thanks in advance =p !

1.

a. resolving the velocity into horizontal and vertical (draw diagram for this)

x velocity = Vcos(7)
y velocity = -Vsin(7) ( note: the y velocity is negative ( the ball is initially falling, this is from the vector diagram and resolution of velocities)

you must convert km/h into m/s ( the SI units)

the rest is standard.

b . y=0, solve quadratic

c. remember tan(theta) = absolute value [(y velocity) / (x velocity)], using the value of t you found in last part. The absolute value ensures you get the acute angle. this is the angle WITH the horizontal!

d. just calculating range, and comparing and giving conclusion.



2. a.

just getting x and y components, eliminating t to get the eqn of the path, and sub in x=20, y=h

b. MAXIUM , any time you see max or min, think calculus!---> calculus, dy/dx=0 , you will have to differentiate sec ^2, a bit of a pain, but get used to it lol.

your best way would be to rewrite sec^2 as [(cos (x) ) ^-1 ]^2= (cosx) ^ -2, then use chain rule. NOTE: thats cos to the power minus 1, NOT INVERSE COS!

c. follows from previous part

d.

you need the x and y velocities when it hits the ground, the speed = sqrt ( (x velocity)^2 + ( y velocity)^2 ) and tan ( theta) = absolute value [(y velocity) / ( x velocity)]

i didnt do the working because i assume u know it as u are doing 4 unit

 

[dishab]

I am also doing the same question no. 14. Cant work out part C. Can someone plz work it out... i am kinda lost...

 

[random-1006]

I am also doing the same question no. 14. Cant work out part C. Can someone plz work it out... i am kinda lost...

let a= x

therefore we have h= 20tanx -5 (cosx)^-2
dh/dx= 20(secx)^2 -5 * -2(cosx)^-3 * -sinx
= 20(secx)^2 -10sinx(cosx)^-3
taking (cosx)^-3 on the bottom and with sinx/cosx= tanx and 1/(cosx)^2 = (secx)^2

dh/dx= 20(secx)^2 -10tanx(secx)^2=0 { for maxium}
10(secx)^2 [ 2-tanx]=0
(secx)^2 =0 ... { no solution} or 2 -tanx=0


ie tanx=2