here
Because the initial direction of velocity is up ( we define as +) , the acceleration is -9.8(against gravity)
(1).... s(y)=u(y) + 1/2a(y)t^2 Right?
Where s is displacement.
Also
(2).... u(y)=usinx
from (2) u(y)=40sinx
since s is displacement then s = -30 [displacement from horizontal(rest) to ground ]
from (1)
-30=40sinxT-4.9T^2 (make t = T so it is clear)
ie 4.9T^2 - 40sinxT -30 = 0
This is a quadratic(which i know u know how to solve Tina)
x= [-b (+ or -) sqrt (b^2 - 4ac) ]/2a
therefore T= [40sinx (+ or -) sqrt (1600sin^2x - 4 * 4.9 * - 30)]/9.8
we take the positive answer (since T is scalar)
Note: I expressed the answer in terms of sinx since the angle is not given. The time taken will vary if the angle is steeper or shallower, hence we cannot find a numerical value)
Now the maximum range can be found since this means x= 45 degrees.
Range means the greatest horizontal distance that can be achieved.
45 degrees is the angle that allows this greatest horizontal distance.
We can use (3) ...
T= [40sinx (+ or -) sqrt (1600sin^2x - 4 * 4.9 * - 30)]/9.8
and for x put 45 degrees to find T
Now deltaX=U(x)T
where T is found from (3) and u(x) is found from u(x)=40cos45
Tell me if this is write mc donalds boy.
By the way Tina to solve Projectile questions u must understand a few concepts.
(1)The velocity throughout the trajectory(the path taken by the projectile which is parabolic) is constantly changing. This trajecotry velocity has Vertical and Horizontal components and the vector some of these independent components is the trajectory's velocity.
(2) There is only one force acting on a projectile when it is in the air and this force is gravity.
(3)Gravity acts in the vertical direction and therefore there is a acceleration in the vertical direction since forces produce accelerations.
(4) All throughout the Projectile the Horizontal Velocity is constant since there is no force to produce an acceleration in the horizontal direction.
(5) When u throw a ball in the air just before it comes down from up , u notice a stop before the turning point, During this stop the velocity is 0 and it has obviously achieved its maximum height.
Therefore at maximum height the velocity is at rest or 0.
(6) the time T is the same time for the vertical and horizontal components since both are normally talking about the time taken in the air.
(7)when they ask what is the velocity when the object strikes the floor they are not talking about the vertical or horizontal velocitys they are talking about the trajectory's velocity which is the vector some of the two components and this velcocity is not 0 because they mean the velocity just before it makes contact.
All this should make u a master in projectile motion.
DOES THIS MAKE SENSE TINA?
