Projectile motion question (1 Viewer)

[catmaster]

Hi everyone
Could sum1 plz help with this question, Its form the cambridge yr 12 3unit textbook question 9c from page 135 (chapter: Projectile motion the equation of path)

The question was
A boy throws a ball with spped V m/s at an angle of 45* to the horizontal

question 9 part (a) is just finding the expression for the displacement components whilst question 9 part b is just deriving the motion's equation to be:

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

by fusing the 2 displacement eqns (i.e. x and y) together

Part c is the part which threw me off, coz i can't prove it without a diagram

(let * be the symbol for degrees, and let @= theta)

c) The boy is now standing on a hill inclined at and angle of @* to the horizontal. He throws the ball at the same angle of elevation of 45* and at the same speed of V m/s. If he can throw the ball 60 metres down the hill, use the result in part (b) to show that

tan@= 1- ((30g Cos@)/(v^2)) = ((60g Cos@)/(v^2)) - 1
and hence show @ = inverse tan (1/3)

in these cases g=10 m/s2 (squared) and y''=-g (for y'' is vertical accelleration)
(for those who have done pm be it maths of physics will know g is gravity, in physics context it is 9.8 m/s2 and in maths g=10m/s2)

Problem is, i can do the maths, just i can't understand what they are asking of me, (which is a very big problem esp in 4u probability :worried
If anyone can draw up a diagram of what im up against, it would be great!!

Much appreciated, thank you everyone

 

[YannY]

Okay dude this is what it is asking you. The angle theta is the hills inclined angle. Now if he stands on top of the hill he would still through at 45* relative to ground, his angle does not change relative with theta. So what it really is asking you to do is to find the triangle length. It should be 1 by 3 by sqrt 10. But you gotta find it in the process. Tan @ = opposite / adjacent