For projectile motion, you know how they sometimes launch an object at an angle from a cliff lets just say 100m of the ground with a ball thrown at some v at 30 degrees to horizontal. Is it safe to calculate time to reach max height. Vy = Uy + Ayt. Then multiply the time by 2 and get range as range = Uxt.
Doing 3U maths, i see in the books they use y= -gt2/2 + vtsin30 + 100 then let y=0 and then figure out time.
The problem im facing is that time in both situations are DIFFERENT. WhICH ONE IS CORRECT?
Time of flight = 2 x Time to reach maximum height ONLY WHEN THE OBJECT LANDS AT THE SAME VERTICAL POINT
What you are finding when you multiply the time to reach max height by 2 is the time taken to reach the initial vertical displacement (i.e., 0) The object still needs to fall another 100m.
What you can do in this case is you can split the motion into 2 seperate parts. It's a tad longer, but makes more sense to some
Firstly, solve for time to reach max height and therefore find time to reach same vertical position.
Then start the second part where the "initial" velocity is equal (but in opposite vertical direction) to the velocity of the object when it was thrown.
Then use Sy = uy*t + 1/2ayt^2
where Sy = -100
Uy = initial vertical velocity (which is equal but opposite to the original vertical velocity)
then use quadratic formula to find t
add this "t" to the Time of flight "t" and you will have your total time of flight
As i said, a tad longer, but makes more sense to some
Essentially, both equations are the same.
y= -gt2/2 + vtsin30 + 100
is the same as saying
Sy = ut + 1/2at^2
Where Sy = -100
-100 = ut + 1/2at^2
Uy = vsin30
-100 = vsin30*t + 1/2at^2
a = -g
-100 = vtsin30 - (gt^2)/2
Take the -100 to the other side to solve using quadratic equation:
- (gt^2)/2 + vtsin30 + 100 = 0
