1999 Q7a iii (1 Viewer)

[Paroissien]

Sorry to bombard you with all these questions.
(all angles in degrees)
A cricket ball leaves the bowler's hand 2m above the ground with a velocity of 30m/s at an angle of 5 below the horizontal. The equations of the motion of the ball are: x(accel) = 0 and y(accel) = -10

i) x = 30tcos(5) and y = -30tsin5 - 5t(squared)

ii) Time at which ball strikes the ground = approx. 0.423 sec

And now,
iii) Calculate the angle at which the ball strikes the gound.
Question doesn't require much work, but I don't understand the theory/method behind. If someone could actually explain their working, that would be awesome.

Cheers

 

[withoutaface]

What you do is find the horizontal and vertical speeds at that time, and then use tan^-1(y'/x')

 

[Paroissien]

What you do is find the horizontal and vertical speeds at that time, and then use tan^-1(y'/x')

And why do you do that? That is the part I got stuck at, I've got the speeds, but why do you use the tan thing?

 

[gordo]

to find the angle, if u have horizontal and vertical component and want the angle
u know tan of the angle is opp over adj
just inverse tan both sides

 

[CrashOveride]

Or if u want u can find the gradient of the tangent at the point where it strikes the ground and get the answer by recalling tan@ = gradient. Then just subtract that from 180