A boy projects a ball vertically upwards with speed 10ms^-1 from a point X, which is 50m above the ground. T seconds after the first ball is projected upwards, the boy drops a second ball from X. Initially the second ball is at rest. The balls collide 25m above the ground. Find the value of T.
-I have tried creating an expression for t in terms of t and T and then I equated expressions of the displacements of both balls in terms of t. I tried to solve for T by solving for t (because my expression for the 2nd ball was t+T), but this didnt work... please help I assume the underlined bit was to suggest furthermore than u=0 for the 2nd ball.
A boy projects a ball vertically upwards with speed 10ms^-1 from a point X, which is 50m above the ground. T seconds after the first ball is projected upwards, the boy drops a second ball from X. Initially the second ball is at rest. The balls collide 25m above the ground. Find the value of T.
-I have tried creating an expression for t in terms of t and T and then I equated expressions of the displacements of both balls in terms of t. I tried to solve for T by solving for t (because my expression for the 2nd ball was t+T), but this didnt work... please help I assume the underlined bit was to suggest furthermore than u=0 for the 2nd ball.
you must have a common origin and well defined direction for positive and negative in this type of problem
My last remark can be ignored as there are no simultaneous equations here. find how long it takes the first ball to get to the collision point. Then take away T seconds from this time. Then start a new equation for the second ball
I don't get why. Let's say the 2nd ball (the one being dropped) is dropped 1 second after the 1st ball is projected- then the 1st ball will have t= (t+1) or (t) and the second ball with have t= (t) or (t-1) respectively? Surely? So why is the first ball having T taken from it in this example?
Plus- when do simultaneous equations ever come into Kinematics M1?
I don't get why. Let's say the 2nd ball (the one being dropped) is dropped 1 second after the 1st ball is projected- then the 1st ball will have t= (t+1) or (t) and the second ball with have t= (t) or (t-1) respectively? Surely? So why is the first ball having T taken from it in this example?
Plus- when do simultaneous equations ever come into Kinematics M1?
FORGET about the second ball !
Find how long it takes the first ball to get 25 metres above ground.
(simultaneous equations in kinematics? Look at the first question of M1. June 2009)
Yeah- the answer is 1.2 seconds(2s.f) but I had to use the negative value from the quadratic to put into the second quadratic. So in total it was a bit tedious and I'm just thinking about why I have to chose the negative root to put into the 2nd quadratic to solve for T because otherwise in an exam i'm stuffed lol.
Yeah- the answer is 1.2 seconds(2s.f) but I had to use the negative value from the quadratic to put into the second quadratic. So in total it was a bit tedious and I'm just thinking about why I have to chose the negative root to put into the 2nd quadratic to solve for T because otherwise in an exam i'm stuffed lol.
I will "borrow" this question so when I have it ready I will post you the solution. (Exactly the same, just different numbers)
I will "borrow" this question so when I have it ready I will post you the solution. (Exactly the same, just different numbers)
What did you mean by common origin for the sim equation question?
It also doesn't make sense why I have to use the negative value of t to solve the quadratic in T and t- because I'm defining t to be the time (in my equation) that it takes ball 1 to go -25 m (upwards is defined as positive) with u=10 ms^-1. so in an exam im stuffed? thanks for the upcoming solution
What did you mean by common origin for the sim equation question?
It also doesn't make sense why I have to use the negative value of t to solve the quadratic in T and t- because I'm defining t to be the time (in my equation) that it takes ball 1 to go -25 m (upwards is defined as positive) with u=10 ms^-1. so in an exam im stuffed? Thank you so much
Your problem was simple (but very interesting) so when I first read it I assumed it is the usual "simultaneous type scenario" which is a lot harder.
So ignore common origin.
Wait for my question and see if this makes sense about the negative.
Your problem was simple (but very interesting) so when I first read it I assumed it is the usual "simultaneous type scenario" which is a lot harder. So ignore common origin Wait for my question and see if this makes sense about the negative.