Projectile Motion Question [HARD] [NO NOOBS] (1 Viewer)

[xXmuffin0manXx]

A hot air balloon is rising obliquely at and anglr of 15 degrees amd at a velocity of 5m/s. A boy who is inside the basket of the balloon throws a ball in the opposite direction of the horizontal motion of the balloon at a velocity at an unknown velocty an angle of 30 degrees. The height of the balloon at the moment the boy throws the ball is 150m. The ball lands 70m away from the moment the ball was thrown.
Ignoring air resistane, what was the initial velocity the ball was thrown at?



Rep to anybody who can get this

 

[xV1P3R]

The height of the balloon at the moment the boy throws the ball is 150m. The ball lands 70m away from the moment the ball was thrown.

So the 70m distance that the ball lands away from the moment the ball was thrown is not the straight line distance, but from what I can gather, the horizontal distance? Also does the ball land on the ground ie. 150m below the moment the ball was thrown.

 

[k02033]

Let u be the speed relative to the ground that the kid throws the ball with.

 

[k02033]

the initial horizontal velocity of the ball relative to the ballon is 13cos(30)+5cos(15) the ballon makes life harder for the kid here

and initial vertical velocity of the ball relative to the ballon is 13sin(30)-5sin(15) the ballon helps the kid in this case

 

[xXmuffin0manXx]

crap.

 

[k02033]

did i get it wrong? whats the right answer?

 

[k02033]

oops i made a typo in one of the lines (where i multiply u into the square root, it should have been 3000u^2 i wrote 3000u, but other than that, the numbers works out nicely... oh and i got lazy and used a=-10ms^-2

 

[Dx_God]

whats so hard about this question? it's a very basic 3u projectile motion question since "The ball lands 70m away from the moment the ball was thrown" and "The height of the balloon at the moment the boy throws the ball is 150m". basically ur first line is just pure rubbish and using y = -g(t^2)/2 + ut/2 + 150 and x = ut(3^1/2)/2 u can replace t with x and sub x= 70, y=0 and do a quadratic and ur done. as for the answer, i can't be bothered doing it and just follow my method and u should be fine with the rest. if u still got problems then u can ask me and i'll do a proper response then.

 

[xXmuffin0manXx]

whats so hard about this question? it's a very basic 3u projectile motion question since "The ball lands 70m away from the moment the ball was thrown" and "The height of the balloon at the moment the boy throws the ball is 150m". basically ur first line is just pure rubbish and using y = -g(t^2)/2 + ut/2 + 150 and x = ut(3^1/2)/2 u can replace t with x and sub x= 70, y=0 and do a quadratic and ur done. as for the answer, i can't be bothered doing it and just follow my method and u should be fine with the rest. if u still got problems then u can ask me and i'll do a proper response then.

im pretty sure it isnt that simple but i may be wrong

 

[Dx_God]

damn just realised i forgot about the horizontal velocity of the balloon and principle of relativity...

 

[Dx_God]

basically ur y function still remains the same as above while ur x function changes. ur x function actually becomes (usin30-5sin15)t due to the fact that the ball is thrown at an opposite direction to the motion of ur balloon. then do the same procedure and u should end up with ur answer

 

[weirdguy99]

So if this is a 3U maths question, do you think it would be HSC Physics material? ie, used in the HSC's?

 

[xXmuffin0manXx]

it's perfectly reasonable and within the syllabus..