A projectile motion question (1 Viewer)

[kooltrainer]

A boy throws a ball with speed Vm/s at an angle of 45 degrees to the horizontal.
a)show the cartesian equation of the path of the ball is y = x - (gx^2)/v^2

b) the boy is now standing on a hill inclined at an angle of a to the horizontal. He throws the ball at the same angle of elevation of 45 degrees and at the same speed V m/s. If he can throw the ball 60 meters down the hill but only 30 meters up the hill, use the result in part (a) to show that :

tan a= 1- (30gcos a)/v^2 = (60gcos a)/v^2 - 1

hence show that tan a = 1/3


note : you dun need to show part (a), its there for your reference.

 

[shaon0]

A boy throws a ball with speed Vm/s at an angle of 45 degrees to the horizontal.
a)show the cartesian equation of the path of the ball is y = x - (gx^2)/v^2

b) the boy is now standing on a hill inclined at an angle of a to the horizontal. He throws the ball at the same angle of elevation of 45 degrees and at the same speed V m/s. If he can throw the ball 60 meters down the hill but only 30 meters up the hill, use the result in part (a) to show that :

tan a= 1- (30gcos a)/v^2 = (60gcos a)/v^2 - 1

hence show that tan a = 1/3


note : you dun need to show part (a), its there for your reference.

I think you need part a) so it is easier to show tan(a)=1/3 in cartesian form.

 

[3unitz]

y = x - (gx^2)/v^2

b) the boy is now standing on a hill inclined at an angle of a to the horizontal. He throws the ball at the same angle of elevation of 45 degrees and at the same speed V m/s. If he can throw the ball 60 meters down the hill but only 30 meters up the hill, use the result in part (a) to show that :

tan a= 1- (30gcos a)/v^2

consider ball being thrown up slope:

sub in point (30cosa, 30sina) into y = x - (gx^2)/v^2:

30sina = 30cosa - (g/v^2) (30cosa)^2

divide through by 30cosa:

tan a = 1 - (30gcosa)/v^2 -----------(1)

= (60gcos a)/v^2 - 1

consider ball being thrown down slope:

same thing but we consider the slope of the hill -a to the horizontal

sub in point (60cosa, -60sina) into y = x - (gx^2)/v^2:

-60sina = 60cosa - (g/v^2) (60cosa)^2

divide through by -60cosa:

tan a = (60gcos a)/v^2 - 1 -----------(2)

hence show that tan a = 1/3

from (2):

tan a = (60gcos a)/v^2 - 1

tan a /2 = (30gcosa)/v^2 - 1/2

- tan a /2 = 1 - (30gcosa)/v^2 -1/2 .... equal to (1)

-tan a /2 = tan a - 1/2

tan a = 1/3